MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 10

Find the length of latus rectum of the parabola

$(a^{2}+b^{2})(x^{2}+y^{2})=(bx+ay-ab)^{2}$

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$\displaystyle \frac{ab}{\sqrt{a^{2}+b^{2}}}$
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$\displaystyle \frac{2ab}{\sqrt{a^{2}+b^{2}}}$
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$a+b$
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$a+b+\dfrac{ab}{\sqrt{ab}}$

Explanation

Alternative Method: The given equation may be written as

$\displaystyle x^{2}+y^{2}=\frac{(bx+ay-ab)^{2}}{(a^{2}+b^{2})}$
or

$\displaystyle \sqrt{x^{2}+y^{2}}=\frac{\left |bx+ay-ab \right |}{\sqrt{a^{2}+b^{2}}}$
or

$\displaystyle \sqrt{(x-0)^{2}+(y-0)^{2}}=\frac{\left |bx+ay-ab \right |}{\sqrt{a^{2}+b^{2}}}$
which is of the form

$SP=PM$
Since distance from focus S to

$\displaystyle (bx+ay-ab=0)=\frac{1}{2}(4\rho )$
or

$\displaystyle \frac{1}{2}(4\rho )=\frac{ab}{\sqrt{a^{2}+b^{2}}}$

$\displaystyle 4\rho =\frac{2ab}{\sqrt{a^{2}+b^{2}}}$

If a hyperbola passes through the foci of the ellipse

$\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

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Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{16} = 1$
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Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{25} = 1$
0%
Focus of the hyperbola is $\displaystyle (5, 0)$
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Focus of the hyperbola is $\displaystyle (5 \sqrt 3, 0)$

Explanation

Given ellipse is,

$\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$
Eccentricity of the ellipse is,

$\displaystyle e_e =\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is,

$(\pm ae_e,0)=(\pm 3 , 0)$
Let eccentricity of the required hyperbola is

$e_h$ and semi major and minor axes are

$a$ and

$b$ , so the equation of hyperbola is,

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough

$(\pm 3,0)\Rightarrow \displaystyle \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2 = 9$
Also given that

$\displaystyle e_e\times e_h = 1$

$\Rightarrow e_h=\cfrac{5}{3}$

$\Rightarrow$

$b^2 =a^2(e_h^2-1)=16$
Hence required hyperbola is,

$\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is,

$(\pm 5,0)$

The locus of the point

$(h,k)$ , if the point

$(\sqrt{3h}, \sqrt{3k + 2})$ lies on the line

$x - y - 1 = 0$ , is a ?

0%
straight line
0%
circle
0%
parabola
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none of these

Explanation

$(\sqrt{3h}, \sqrt{3k + 1})$ lines on the line

$x - y - 1 = 0$

$\Rightarrow (\sqrt{3h})^2 = (\sqrt{3k + 2} + 1)^2$

$\Rightarrow 3h = 3k + 2 + 1 + 2 \sqrt{3 k + 2}$

$\Rightarrow 3^2 (h - k - 1)^2 = 2^2 (\sqrt{3k + 2})^2$

$\Rightarrow 9(h^2 + k^2 + 1 - 2hk - 2h + 2k) = 4(3k + 2)$

$\Rightarrow 9(x^2 + y^2) - 18xy - 18x + 6y + 1 = 0$
Now

$h^2 = ab$ and

$\Delta \neq 0$
Therefore, locus is a parabola.

A circle touches the

$x$ -axis and also touches the circle with centre

$(0, 3)$ and radius

$2$ . The locus of the centre of the circle is -

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a circle
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an ellipse
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a parabola
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a hyperbola

Explanation

Let

$C_1(h, k)$ be the center of the circle.

Circle touches the

$x$ -axis then its radius is

$r_1 = k$ .

Also circle touches the circle with centre

$C_2(0, 3)$ and radius

$r_2 = 2$ .

$\therefore |C_1 C_2| = r_1 + r_2$

$\Rightarrow \sqrt{(h - 0)^2 + (k - 3)^2} = |k+2|$

Squaring

$h^2 - 10k + 5 = 0$

$\Rightarrow$ Locus is

$x^2 - 10y + 5 = 0$ , which is parabola.

A point

$(\alpha, \beta)$ lies on a circle

$x^2+y^2=1$ , then locus of the point

$(3\alpha +2\beta)$ is a

$/$ an.

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Straight line
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Ellipse
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Parabola
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None of these

Explanation

Point will be

$(3\alpha ,2\beta )$ not

$( 3\alpha +2\beta )$

Now

$x^{2}+y^{2}=1$

Radium is

$1$ unit,hence parametric co - ordinate is

$(\alpha ,\beta )= (1\cos\theta ,1\sin\theta )=(\cos\theta , \sin\theta )$

Hence

Point is

$(3\cos\theta ,2\sin\theta )$

Hence

$(x,y)= (3\cos\theta ,2\sin\theta )$

$x=3\cos\theta$

$\Rightarrow \dfrac{x}{3}\cos\theta$ ...(i)

$y=2\sin\theta$

$\Rightarrow \dfrac{y}{2} = \sin \theta$ ...(ii)

$(i)^{2} + (ii)^{2}$

$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} \cos^{2}\theta + \sin^{2} \theta$

$\dfrac{x^{2}}{9}+ \dfrac{y^{2}}{4} = 1$

which is equation of ellipse

Find the equation of the circle that passes through the points

$(0,6),(0,0)$ and

$(8,0)$

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${(x-4)}^{2}+{(y-3)}^{2}=25$
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$(x+4)^2+(y+3)^2=25$
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$(x-3)^2+(y-4)^2=25$
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$(x-4)^2+(y-3)^2=36$

Explanation

Let the equation of the general form of the required circle be

$x^2+y^2+2gx+2fy+c=0$ ................(1)

According to the problem, the above equation of the circle passes through the points $(0, 6), (0, 0)$ and $(8, 0)$ . Therefore,

$36 + 12f + c = 0$ ………. (2)

$c = 0$ ……………. (3)

$64 + 16g + c = 0$ ……………. (4)

Putting $c = 0$ in (2), we obtain $f = -3$ . Similarly put $c = 0$ in (4), we obtain $g=-4$

Substituting the values of $g, f$ and $c$ in (1), we obtain the equation of the required circle as:

$x^2+y^2+2(-4)x+2(-2)y+0=0$ that is

$x^2+y^2-8x-4y+0=0$ can be rewritten as

$x^2+y^2-8x-4y+16+9=0+16+9$

$(x-4)^2+(y-3)^2=25$

Therefore, the equation of circle is $(x-4)^2+(y-3)^2=25$ .

A circle and a parabola intersect at four points

$(x_1 , y_1), (x_2 , y_2), (x_3 , y_3)$ and

$(x_4 , y_4)$ . Then

$y_1 + y_2 + y_3 + y_4$ is equal to

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$4$
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$3/2$
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$2$
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$0$

Explanation

A circle and a parabola intersect at four points

$\left( { x }_{ 1 },{ y }_{ 1 } \right) ;\left( { x }_{ 2 },{ y }_{ 2 } \right) ;\left( { x }_{ 3 },{ y }_{ 3 } \right) ;\left( { x }_{ 4 },{ y }_{ 4 } \right)$ .

Then

${ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }$ is equal to :

${ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }={ 2at }_{ 1 }+{ 2at }_{ 2 }+{ 2at }_{ 3 }+{ 2at }_{ 4 }=0$

A Iight ray gets reflected

from the

$x = -2$ . If the reflected touches the circle

$x^2 + y^2 = 4$ and point of incident is

$(-2,-4)$ , then equation of incident ray is

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$4y+3x+22=0$
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$3y+4x+20=0$
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$4y+2x+20=0$
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$y+x+6=0$

Find the equation of the circle with center at

$(-3,5)$ and passes through the point

$(5,-1)$

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${(x+3)}^{2}+{(y-5)}^{2}=100$
0%
$(x-3)^2+(y-5)^2$
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$(x+3)^2+(y-5)^2$
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None of the above

Explanation

Here we know that

$(h,k)=(-3,5)$ but we are not given the radius.

However, we can find the radius by using the distance formula.

Therefore,

$r = \sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}$

$r = \sqrt {(5-(-3) )^2 + (-1-5)^2}$

$r = \sqrt {(8 )^2 + (6)^2}$

$r = \sqrt {64+36}$

$r = \sqrt {100}$

$r=10$

An equation of the circle with center

$(h,k)$ and radius

$r$ is

$(x-h)^{ 2 }+(y-k)^{ 2 }=r^{ 2 }$

Substitute the values in the equation of circle as follows:

$(x-(-3))^{ 2 }+(y-5)^{ 2 }=10^{ 2 }$

i.e.,

$(x+3)^{ 2 }+(y-5)^{ 2 }=100$

Therefore, the equation of the circle is

$(x+3)^{ 2 }+(y-5)^{ 2 }=100$ .

Normals

$AA,A{A}_{1}$ and

$A{A}_{2}$ are drawn to the parabola

${ y }^{ 2 }=8x$ from the point

$A(h,0)$ . If triangle

$O{A}_{1}{A}_{2}$ is equilateral , then the possible value of

$h$ is

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$26$
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$24$
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$28$
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None of these

Consider a parabola

$P$ which touches

$y = 0$ at

$(1, 0)$ and

$x = 0$ at

$(0, 2)$ , then latus rectum of

$P$ is,

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$\dfrac {9}{5\sqrt {5}}$
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$\dfrac {16\sqrt {5}}{5}$
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$\dfrac {16}{25}$
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$\dfrac {8}{25}$

Explanation

Let

$(a,b)$ be the focus and

$y=m{x}$ be directrix

$\implies \bigg(\dfrac{m{x}-y}{\sqrt{1+m^{2}}}\bigg)^{2}=(x-a)^{2}+(y-b)^{2}$

Differentiating on both sides

$\dfrac{(m{x}-y)(m-\dfrac{d{y}}{d{x}})}{1+m^{2}}=(x-a)+(y-b)\dfrac{d{y}}{d{x}}$

$x$ axis is tangent at

$(1,0)$

$\dfrac{m(m)}{1+m^{2}}=1-a\implies a=\dfrac{1}{1+m^{2}}$

$y$ axis is tangent at

$(0,2)$

$\dfrac{-2}{1+m^{2}}=b-2\implies b=\dfrac{2{m}^{2}}{1+m^{2}}$

$(1,0)$ lies on parabola

$\dfrac{(m)^{2}}{1+m^{2}}=(1-a)^{2}+b^{2}$

Substituting

$a$ and

$b$ values

$\implies (4{m^2}-1)(m^{2})=0\implies m=0,\pm \dfrac{1}{2}$

For

$m=0$ we get

$a=1,b=0$ which means that the directrix cuts the parabola which is not possible so

$m=\pm \dfrac{1}{2}$

$\implies a=\dfrac{4}{5},b=\dfrac{2}{5}$

So the focus is

$\bigg(\dfrac{4}{5},\dfrac{2}{5}\bigg)$

the directrix is

$2{y}+x=0$

The axis will be

$2{x}-y+c=0$

Focus lies on the axis

$\implies c=-\dfrac{6}{5}$

the point of intersection of axis and directrix is

$\bigg(\dfrac{12}{25},\dfrac{-6}{25}\bigg)$

Vertex will be

$\bigg(\dfrac{16}{25},\dfrac{2}{25}\bigg)$

Latus Rectum will be

$4$ times the distance between focus and vertex which is equal to

$\dfrac{16\sqrt{5}}{5}$

Hence option

$B$ is the answer.

A circle touches the y-axis at

$(0, 2)$ and has an intercept of

$4$ units on the positive side of the x-axis. Then the equation of the circle is?

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$x^2+y^2-4(\sqrt{2}x+y)+4=0$
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$x^2+y^2-4(x+\sqrt{2}y)+4=0$
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$x^2+y^2-2(\sqrt{2}x+y)+4=0$
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none of these

Explanation

Let $s: x^2 + y^2 + 2gx + 2fy +c = 0$ be the equation of the circle

$S(0,2) = 0$

$\implies 4 + 4f + c =0 -eq.1$

Intercept on x-axis is given by

$2\sqrt{g^2 - c} = 4$

$g^2 – c = 4 -eq.2$

Since x = 0 is a tangent

Radius = perpendicular distance from the center to tangent

$$\sqrt {g^2 + f^2 - c} = \dfrac{|-g|}{\sqrt{1 + 0}$$

$g^2 + f^2 - c = g^2$

$f^2 = c$

Substituting in eq.1

$4 + 4f + f^2 =0 \implies (f+2)^2 = 0 \implies f = -2$

$\implies c = 4$

From eq.2

$g^2 – 4 = 4$

$g = \pm 2 \sqrt{2}$

Equation of circle is

$x^2 +y^2 \pm 4\sqrt 2 x – 4y + 4 = 0$

$\implies x^2 +y^2 – 4(\sqrt 2 x + y) + 4 = 0$

Let

$S$ be the focus of

${ y }^{ 2 }=4x$ and a point

$P$ be moving on the curve such that its abscissa is increasing at the rate of

$4 units/s$ . Then the rate of increase of the projection of

$SP$ on

$x+y=1$ when

$P$ is at

$(4,4)$ is

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$\sqrt { 2 }$
0%
$-1$
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$-\sqrt { 2 }$
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$-3\sqrt { 2 }$

O is the centre of a circle of diameter

$4$ cm and OABC is a square, if the shaded area is

$\displaystyle\frac{1}{3}$ area of the square, then the side of the square is __________.

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$\pi\sqrt{3}$ cm
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$\sqrt{3\pi}$ cm
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$3\sqrt{\pi}$ cm
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$3\pi$ cm

Explanation

$O$ is the center of the circle and the diameter is

$4$ units.

Also area of the shaded region

$=\cfrac { 1 }{ 3 }$ area of the square

Area of the shaded region

$=\pi { r }^{ 2 }\times \cfrac { 90 }{ 360 } \\ =\pi \left( 4 \right) \left( \cfrac { 1 }{ 4 } \right) \\ =\pi$

Area of the square

$={ a }^{ 2 }=\pi \times 3=3\pi$

$\therefore a=\sqrt { 3\pi }$ cm

1141297_727707_ans_7110365d9b7842f29fa116a0987898df.png

The ratio of the areas of the triangle

$PQS$ and

$PQR$ is :

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$1 : \sqrt2$
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$1 : 2$
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$1 : 4$
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$1 : 8$

Explanation

Equation of parabola is

$y^2=8x$

Equation of circle is

$x^2+y^2=9$

To find point of intersection of both, put equation of parabola in equation of circle.

$\therefore x^2+8x=9$

$\therefore x^{ 2 }+8x-9=0$

$\therefore x^{ 2 }+9x-x-9=0$

$\therefore x\left( x+9 \right) -1\left( x+9 \right) =0$

$\therefore \left( x+9 \right) \left( x-1 \right) =0$

$\therefore x=-9$ and

$x=1$

But,

$x\neq -9$ as

$y^2$ cannot be negative

$\therefore x=1$

From equation of parabola,

${ y }^{ 2 }=8$

$\therefore { y }=\pm 2\sqrt { 2 }$

Thus, points of intersection are

$P\left( 1,2\sqrt { 2 } \right)$ and

$Q\left( 1,-2\sqrt { 2 } \right)$

Consider tangent to the circle which intersects x axis at point R.

Equation of tangent is,

$x{ x }_{ 1 }+y{ y }_{ 1 }=9$

$\therefore x\left( 1 \right) +y\left( 2\sqrt { 2 } \right) =9$

$\therefore x+2\sqrt { 2 } y=9$

When

$y=0$ ,

$x=9$

Thus, coordinates of point R are

$R\left( 9,0 \right)$

Consider tangent to the parabola which intersects x axis at point S.

Equation of tangent is,

$y{ y }_{ 1 }=4a\left( x+{ x }_{ 1 } \right)$

$\therefore y\left( 2\sqrt { 2 } \right) =8\left( x+1 \right)$

When

$y=0$ ,

$x=-1$

Thus, coordinates of point S are

$R\left( -1,0 \right)$

Area of triangle PQS is,

$A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times PQ\times SM$

$\therefore A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 2$

$\therefore A\left( \triangle PQS \right) =4\sqrt { 2 }$

Area of triangle PQR is,

$A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times PQ\times RM$

$\therefore A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 8$

$\therefore A\left( \triangle PQR \right) =16\sqrt { 2 }$

$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =\dfrac { 4\sqrt { 2 } }{ 16\sqrt { 2 } }$

$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =1:4$

1973247_874922_ans_2efa3d16990d43a3ba04c0ccb23185cd.png

If eccentricity of both ellipses are same, then their eccentricity is

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$2-\sqrt { 3 }$
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$\sqrt { 2 } -1$
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$\cfrac { 3-\sqrt { 5 } }{ 2 }$
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$\cfrac { \sqrt { 5 } -1 }{ 2 } \quad$

An endless inextensible string of length

$15$ m passes around the pins, A & B which are

$5$ m apart. This string is always kept tight and a small ring, R of negligible dimensions, inserted in this string is made to move in a path keeping all segments RA, AB, RB tight (as mentioned earlier). The ring traces a path, given by conic C, then.

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Conic C is an ellipse with eccentricity $\displaystyle\frac{1}{2}$
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Conic C is an hyperbola with eccentricity $2$
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Conic C is an ellipse with eccentricity $\displaystyle\frac{2}{3}$
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Conic C is a hyperbola with eccentricity $\displaystyle\frac{3}{2}$

$S_{1}$ and

$S_{2}$ are the foci of an ellipse of major axis of length 10 units, and P is any point on the ellipse such that the perimeter or triangle

$PS_{1}S_{2}$ isThen the eccentricity of the ellipse is

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0.5
0%
0.25
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0.28
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0.75

$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad$

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$(1,1)$
0%
$(0,2)$
0%
$(2,0)$
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$None\quad of\quad these$

Explanation

Given Hyperbola:

$x^{2}+4y^{2}+6xy+8x-2y+7=0$

Centre: Point of intersection of asymptotes of hyperbola.

Now finding Asymptotes of given equation, taking equation of asymptote

$y=mx+c$ by replacing

$x\rightarrow 1, y\rightarrow m$ in

$\phi_n(m)$

When

$n=2$ ,

$\phi_{2}(m)=1+4m^{2}+6m$

$\phi_{1}(m)=8-2m$

$\phi_{0}(m)=7$

$\phi_{2}^{1}(m)=8m+6$

$\phi_{1}(m)=8-2m$

Putting

$\phi_{2}(m)=0$

$\Rightarrow 1+6m+4m^{2}=0$

$\Rightarrow m=\cfrac{-6\pm \sqrt {36-16}}{2(4)}$

$\Rightarrow m=\cfrac{-6\pm\sqrt 20}{2(4)}$

$\Rightarrow m=\cfrac{-3\pm \sqrt 5}{4}$

So, m

$=\cfrac{-3+\sqrt 5}{4}, \cfrac{-3-\sqrt 5}{4}$

Value of

$c$ , when

$m$ is different

$c= \cfrac{-\phi_{1}(m)}{\phi_{2}^{'}(m)}=\cfrac{8-2m}{8m+6}$

For

$m=\cfrac{-3+\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3+\sqrt 5}{4})}{8\cfrac{-3+\sqrt 5}{4})+6}=\cfrac{3-\sqrt 5+16}{-6+2\sqrt 5+6}=\cfrac{19\sqrt 5-5}{10}$

For

$m=\cfrac{-3-\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3-\sqrt 5}{4})}{8\cfrac{-3-\sqrt 5}{4})+6}=\cfrac{19+15}{-2 \sqrt 5}=\cfrac{-(19\sqrt 5+5)}{10}$

Equation of Asymptotes :

$y=\cfrac{-3+\sqrt 5}{4}x +\cfrac{19\sqrt 5-5}{10}$ &

$y=\cfrac{-3-\sqrt 5}{4}x-(\cfrac{5+19\sqrt 5}{10})$

On solving them for x & y, putting LHS-RHS

$\Rightarrow 0=\cfrac{\sqrt 5}{2}x+\cfrac{19\sqrt 5}{5} \Rightarrow x=\cfrac{-38}{5}$

Now putting values of x in Asymptotes equation, we get

$y=\cfrac{-3-\sqrt 5(-19)}{10}+\cfrac{+5+19\sqrt 5}{10}=\cfrac{+52}{10}$

Centre

$(\cfrac{-38}{5}, \cfrac{+52}{10})$ .

'O' is the vertex of the parabola

${ y }^{ 2 }=8x$ and L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H, then the length of the double ordinate through H is

$\lambda \sqrt { 5 }$ where

$\lambda$ is equal to

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$2$
0%
$4$
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$6$
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$8$

Which of the following equations does not represent a hyperbola?

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$xy = 4$
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$\dfrac{1} {x^2} + \dfrac{1} {y^2} = \dfrac{1} {4}$
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$x^2 - xy + y^2 = 4$
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$x^2 - 4xy + 3y^2 = 1$

The set of points

$(x, y)$ whose distance from the line

$y = 2x + 2$ is the same as the distance from

$(2, 0)$ is a parabola. This parabola is congruent to the parabola in standard form

$y = Kx^{2}$ for some

$K$ which is equal to

0%
$\dfrac {\sqrt {5}}{12}$
0%
$\dfrac {\sqrt {5}}{4}$
0%
$\dfrac {4}{\sqrt {5}}$
0%
$\dfrac {12}{\sqrt {5}}$

Explanation

Distance from

$(2, 0)$ to

$2x - y + 2 = 0$ is equal to

$\dfrac {6}{\sqrt {5}} =$ semi latus rectum

$\therefore$ length of latus rectum is

$\dfrac {12}{\sqrt {5}}$

$\therefore x^{2} = \dfrac {1}{K}y \Rightarrow \dfrac {1}{K} = \dfrac {12}{\sqrt {5}}$

$\Rightarrow K = \dfrac {\sqrt {5}}{12}$ .

From the point

$A\left(0,3\right)$ on the circle

${x}^{2}-4x+{\left(y-3\right)}^{2}=0$ a chord

$AB$ is drawn and extended to a point

$M$ such that

$AM=2AB$ .The locus is

0%
${x}^{2}+{y}^{2}-8x-6y+9=0$
0%
${x}^{2}+{y}^{2}+8x-6y+9=0$
0%
${x}^{2}+{y}^{2}-8x+6y+9=0$
0%
${x}^{2}+{y}^{2}+8x+6y+9=0$

Explanation

${x}^{2}-4x+{\left(y-3\right)}^{2}=0$
Add

$4$ both sides of the above equation, we get

$\Rightarrow {x}^{2}-4x+4+{\left(y-3\right)}^{2}=4$

$\Rightarrow {\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=4$
Take

$B=\left(2+2\cos\theta,3+2\sin\theta\right)$

$\therefore 2+2\cos\theta=\dfrac{x+0}{2} , 3+2\sin\theta=\dfrac{y+3}{2}$
or

$2\cos\theta=\dfrac{x}{2}-2, 2\sin\theta=\dfrac{y+3}{2}-3$
or

$2\cos\theta=\dfrac{x-4}{2}, 2\sin\theta=\dfrac{y-3}{2}$
Eliminating

$\theta$ we get

${\left(2\cos\theta\right)}^{2}+{\left(2\sin\theta\right)}^{2}={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$

$4={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$

${\left(x-4\right)}^{2}+{\left(y-3\right)}^{2}=16$
On simplifying , we get

${x}^{2}+{y}^{2}-8x-6y+9=0$

The line

$2x + y = 1$ is tangent to the hyperbola

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ . If this line passes through the point of intersection of the nearest directrix and the x-axis, then eccentricity of the hyperbola.

0%
1
0%
2
0%
3
0%
4

The equation of the circle having the lines

$x^2 + 2xy + 3x + 6y = 0$ as its normals and having size just sufficient to contain the circle

$x (x - 4) + y(y - 3) = 0$ is

0%
$x^2 + y^2 + 3x - 6y - 40 = 0$
0%
$x^2 + y^2 + 6x - 3y - 45 = 0$
0%
$x^2 + y^2 + 8x + 4y - 20 = 0$
0%
$x^2 + y^2 + 4x + 8y + 20 =0$

Explanation

Pair of lines:

$x^{2}+2xy+3x+6y=0$

$\Rightarrow x(x+2y)+3(x+2y)=0$

$\Rightarrow (x+2y)(x+3)=0$

$\Rightarrow x+2y=0$

and

$x+3=0$

Are the two normals and their point of intersection must be the centre of the circle.

$\Rightarrow x=-3$

and

$y=\dfrac{-x}{2}=\dfrac{3}{2}$

$\Rightarrow (-3,\dfrac{3}{2})$ is the centre of required circle.

(

$\because x(x-4)+y(y-3)=0$ is the diametric form of the circle. Hence,

$(0,0)$ and

$(4,3)$ are the diametric end points.)

$\Rightarrow$ Centre

$\rightarrow (\dfrac{0+4}{2},\dfrac{0+3}{2})\rightarrow (2,\dfrac{3}{2})$

radius

$=\dfrac{1}{2}(\sqrt{16+9})$

$=\dfrac{5}{2}$

The required circle with centre

$(-3,\dfrac{3}{2})$ is just sufficient to contain the circle

$x(x-4)+y(y-3)=0$

$\therefore$ radius of required circle

=distance between

$(-3,\dfrac{3}{2})$ and

$(2,\dfrac{3}{2})$

$+$ radius of given circle.

$=\sqrt{(-3-2)^{2}+0}+\dfrac{5}{2}$

$=5+\dfrac{5}{2}=\dfrac{15}{2}$

$\therefore$ Equation of required circle:

$\Rightarrow (x+3)^{2}+(y-\dfrac{3}{2})^{2}=\dfrac{225}{4}$

$\Rightarrow x^{2}+y^{2}+9+\dfrac{9}{4}+6x-3y=\dfrac{225}{4}$

$\Rightarrow x^{2}+y^{2}+6x-3y+9+\dfrac{9}{4}-\dfrac{225}{4}=0$

$\Rightarrow x^{2}+y^{2}+6x-3y+9-\dfrac{216}{4}=0$

$\Rightarrow x^{2}+y^{2}+6x-3y+9-54=0$

$\Rightarrow x^{2}+y^{2}+6x-3y-45=0$

$\therefore B)$ Answer.

996790_1032066_ans_90bcf08044d74ce4802931eb0ded2397.JPG

$LL^1$ is the latus rectum of an ellipse and

$\Delta S^1LL^1$ is an equilateral triangle. Then

$e=?$

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$\dfrac{1}{\sqrt{2}}$
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$\dfrac{1}{\sqrt{3}}$
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$\dfrac{1}{\sqrt{5}}$
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$\sqrt{\dfrac{2}{3}}$

Explanation

Let equation of the ellipse is

$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ b^{ 2 } } =1$

Refer the figure. By distance formula,

${ S }^{ ' }L=\sqrt { { \left( ae-\left( -ae \right) \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$

$\therefore { S }^{ ' }L=\sqrt { { \left( 2ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$

$\therefore { S }^{ ' }L=\sqrt { 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } }$

$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }=4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } }$ (1)

Similarly,

$L{ L }^{ ' }=\sqrt { { \left( ae-ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } -\left( -\frac { { b }^{ 2 } }{ a } \right) \right) }^{ 2 } }$

$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$

$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { 2{ b }^{ 2 } }{ a } \right) }^{ 2 } }$

$\therefore L{ L }^{ ' }=\sqrt { { \frac { 4{ b }^{ 4 } }{ { a }^{ 2 } } } }$

$\therefore { \left( L{ L }^{ ' } \right) }^{ 2 }=\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } }$ (2)

Now, given

$\triangle { S }^{ ' }LL^{ ' }$ is equilateral triangle

$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }={ \left( L{ L }^{ ' } \right) }^{ 2 }$

$\therefore 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } =\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } }$

$\therefore 4{ a }^{ 2 }{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 2 } }$

$\therefore 4{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 4 } }$ (3)

Now,

$\frac { { b }^{ 2 } }{ { a }^{ 2 } } =1-{ e }^{ 2 }$

$\therefore { \left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) }^{ 2 }={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$

$\therefore \frac { { b }^{ 4 } }{ { a }^{ 4 } } ={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$ (4)

From equations (3) and (4), we can write,

$4{ e }^{ 2 }=3{ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$

$\therefore 4{ e }^{ 2 }=3\left( 1-2{ e }^{ 2 }+{ e }^{ 4 } \right)$

$\therefore 4{ e }^{ 2 }=3-6{ e }^{ 2 }+3{ e }^{ 4 }$

$\therefore 3{ e }^{ 4 }-10{ e }^{ 2 }+3=0$

$\therefore 3{ e }^{ 4 }-9{ e }^{ 2 }-{ e }^{ 2 }+3=0$

$\therefore 3{ e }^{ 2 }\left( { e }^{ 2 }-3 \right) -1\left( { e }^{ 2 }-3 \right) =0$

$\therefore \left( 3{ e }^{ 2 }-1 \right) \left( { e }^{ 2 }-3 \right) =0$

$\therefore { e }^{ 2 }=\frac { 1 }{ 3 }$ or

${ e }^{ 2 }=3$

$\therefore { e }=\frac { 1 }{ \sqrt { 3 } }$ or

${ e }=\sqrt { 3 }$

But

${ e }<1$

$\therefore { e }\neq \sqrt { 3 }$

$\therefore { e }=\frac { 1 }{ \sqrt { 3 } }$

1946415_1082283_ans_a6c838f567cd4a21a982152f00f47135.png

Circles are drawn on chords of the rectangular hyperbola

$xy=4$ parallel to the line

$y=x$ as diameters.All such circles pass through two fixed points whose coordinates are

0%
$\left(2,2\right)$
0%
$\left(2,-2\right)$
0%
$\left(-2,2\right)$
0%

$\left(-2,-2\right)$

Explanation

Given:Rectangular hyperbola

$xy=4={c}^{2}$

$\Rightarrow\,{c}^{2}=4$

$\Rightarrow\,c=2$

Let

$P$ and

$Q$ be the end points on the Rectangular hyperbola where

$P\left(2{t}_{1},\dfrac{2}{{t}_{1}}\right)$ and

$Q\left(2{t}_{2},\dfrac{2}{{t}_{2}}\right)$

Using the end points of diameter the equation of the circle

$C:\left(x-2{t}_{1}\right)\left(x-2{t}_{2}\right)+\left(y-\dfrac{2}{{t}_{1}}\right)\left(y-\dfrac{2}{{t}_{2}}\right)=1$ ..........

$(1)$

Now,Slope of the

$PQ=\dfrac{\dfrac{2}{{t}_{2}}-\dfrac{2}{{t}_{1}}}{2{t}_{2}-2{t}_{1}}$

$=\dfrac{2\left(\dfrac{1}{{t}_{2}}-\dfrac{1}{{t}_{1}}\right)}{2\left({t}_{2}-{t}_{1}\right)}$

$=\dfrac{\dfrac{{t}_{1}-{t}_{2}}{{t}_{1}{t}_{2}}}{\left({t}_{2}-{t}_{1}\right)}$

$=\dfrac{-1}{{t}_{1}{t}_{2}}$

Hence

$PQ$ is the diameter for circle and it is parallel to the line

$y=x$

Slope of

$PQ=$ Slope of the line

$y=x$

$\Rightarrow\,\dfrac{-1}{{t}_{1}{t}_{2}}=1$

$\Rightarrow\,{t}_{1}{t}_{2}=-1$

$(1)\Rightarrow\,\left(x-2{t}_{1}\right)\left(x-2{t}_{2}\right)+\left(y-\dfrac{2}{{t}_{1}}\right)\left(y-\dfrac{2}{{t}_{2}}\right)=1$

$\Rightarrow\,x\left(x-2{t}_{2}\right)-2{t}_{1}\left(x-2{t}_{2}\right)+y\left(y-\dfrac{2}{{t}_{2}}\right)-\dfrac{2}{{t}_{1}}\left(y-\dfrac{2}{{t}_{2}}\right)=1$

$\Rightarrow\,{x}^{2}-2x{t}_{2}-2x{t}_{1}+4{t}_{1}{t}_{2}+{y}^{2}-\dfrac{2y}{{t}_{2}}-\dfrac{2y}{{t}_{1}}+\dfrac{4}{{t}_{1}{t}_{2}}=1$

$\Rightarrow\,{x}^{2}-2x{t}_{2}-2x{t}_{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t}_{2}}-\dfrac{2y}{{t}_{1}}+\dfrac{4}{\times -1}=1$ using

${t}_{1}{t}_{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t}_{2}+{t}_{1}\right)-4-2y\left(\dfrac{1}{{t}_{2}}+\dfrac{1}{{t}_{1}}\right)-4=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)-2y\left(\dfrac{{t}_{1}+{t}_{2}}{{t}_{1}{t}_{2}}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)-2y\left(\dfrac{{t}_{1}+{t}_{2}}{-1}\right)=1$ using

${t}_{1}{t}_{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t}_{2}+{t}_{1}\right)+2y\left({t}_{1}+{t}_{2}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t}_{2}+{t}_{1}\right)=1$ is of the form

$C+\lambda\,L$

where

$C={x}^{2}+{y}^{2}-8=0$ is the equation of a circle.

and

$L=2y-2x=0$ is the equation of a line.

$\Rightarrow\,y-x=0$ or

$x=y$

Substituting

$x=y$ in the equation

${x}^{2}+{y}^{2}-8=0$ we get

$\Rightarrow\,2{x}^{2}-8=0$

$\Rightarrow\,2\left({x}^{2}-4\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$

$\therefore\,x=2,-2$

$\Rightarrow\,y=2,-2$ since

$x=y$

Hence the coordinates of the fixed points are

$\left(2,2\right)$ and

$\left(-2,-2\right)$

Consider
the set of hyperbola

$xy = {\text{ }}K,{\text{ K}} \in {\text{R,}}$ let

${e_1}$ be eccentricity
when

$K = \sqrt {2017}$ and

${e_2}$ be the
eccentricity when

$K = \sqrt {2018}$ , then

${e_1} - {e_2}$ is equal to

0%
-1
0%
0
0%
2
0%
1

The centre of a circle is

$(2, -3)$ and the circumference is

$10\pi$ . Then, the equation of the circle is

0%
${x}^{2}+{y}^{2}+4x+6y+12=0$
0%
${x}^{2}+{y}^{2}-4x+6y+12=0$
0%
${x}^{2}+{y}^{2}-4x+6y-12=0$
0%
${x}^{2}+{y}^{2}-4x-6y-12=0$

Explanation

The circumference of the circle is given as $10\pi$ .

$C = 2\pi r$

$10\pi = 2\pi r$

$r = \frac{{10\pi }}{{2\pi }}$

$r = 5$

The general form of the equation is,

${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$

The center of the circle is given as $\left( {h,k} \right) = \left( {2, - 3} \right)$

${\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {\left( 5 \right)^2}$

${\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 25$

${x^2} + 4 - 4x + {y^2} + 9 + 6y = 25$

${x^2} + {y^2} - 4x + 6y + 13 = 25$

${x^2} + {y^2} - 4x + 6y - 25 + 13 = 0$

${x^2} + {y^2} - 4x + 6y - 12 = 0$

Therefore, the equation of the circle is,

${x^2} + {y^2} - 4x + 6y - 12 = 0$

A conic

$C$ passes through the points

$(2,4)$ and is such that the segment of any of its tangents at any point contained between the co-ordinate axis is biscected at the point of tangency. Let

$S$ denotes circle described on the foci

${F_1}$ and

${F_2}$ of the conic

$C$ as diameter.
Equation of the circle

$S$ is

0%
${x^2} + {y^2} = 16$
0%
${x^2} + {y^2} = 8$
0%
${x^2} + {y^2} = 32$
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${x^2} + {y^2} = 4$

The centre of a circle passing through the point

$(0,0),(1,0)$ and touching the circle

$x^{2}+y^{2}=9$ is ?

0%
$\left(\dfrac {3}{2},\dfrac {1}{2}\right)$
0%
$\left(\dfrac {1}{2},\dfrac {3}{2}\right)$
0%
$\left(\dfrac {1}{2},\dfrac {1}{2}\right)$
0%
None of these

Explanation

Let the circle be

$x^{2}+y^{2}+2gx+2fy+c=0$

$\because$ It passes through

$(0,0)$

$\Rightarrow 0+c=0$

$\Rightarrow c=0$

It also passes through

$(1,0)$

$\Rightarrow 1+0+2g+0+c=0$

$\Rightarrow g=\dfrac{-c-1}{2}=\dfrac{-1}{2}$

$\therefore$ Circle

$\rightarrow x^{2}+y^{2}-x+2fy=0$

Centre

$\rightarrow (\dfrac{1}{2},-f)$

Other circle :

$x^{2}+y^{2}=9$

Centre \rightarrow (0,0)$$

Radius

$=3$

$$\because Circles touches internally.

$\therefore$ Difference between radius = Distance between centres.

Radius of smaller circle

$=\sqrt{g^{2}+f^{2}-0}=\sqrt{\dfrac{1}{4}+f^{2}}$

$\Rightarrow 3-\sqrt{\dfrac{1}{4}+f^{2}}=\sqrt{(-g-0)^{2}+(-f-0)^{2}}$

$\Rightarrow 3-\sqrt{\dfrac{1 }{4}+f^{2}}=\sqrt{\dfrac{1}{4}+f^{2}}$

$\Rightarrow 2\sqrt{\dfrac{1}{4}+f^{2}}=3$

$\Rightarrow \dfrac{1}{4}+f^{2}=\dfrac{9}{4}$

$\Rightarrow f^{2}=\dfrac{8}{4}=2$

$\Rightarrow f=\pm \sqrt{2}$

Hence centre

$\rightarrow (-g,-f)$

$\Rightarrow(\dfrac{1}{2},\pm\sqrt{2})$

$\therefore$ None of the options is correct.

1078662_1039406_ans_03904100fdfd496bbb463c341c7f2acf.PNG

If the centroid of an equilateral triangle

$(1,1)$ and its one vertex is

$(-1,2)$ , then the equation of the circumcircle is

0%
${x^2} + {y^2} - 2x - 2y - 3 = 0$
0%
${x^2} + {y^2} + 2x - 2y - 3 = 0$
0%
${x^2} + {y^2} + 2x + 2y - 3 = 0$
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${(x + 2)^2} + {y^2} = 5$

The focus of extremities of the latus rectum of the family of the ellipse

${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}{\text{ is }}\left( {b \in R} \right)$

0%
${x^2} - ay = {a^3}$
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${x^2} - ay - {e^2}$
0%
${x^2} \pm ay = {a^2}$
0%
${x^2} + ay - {b^2}$

The equation of the circle touches y axis and having radius

$2$ units and centre is

$(-2, -3)$ ?

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$x^2+y^2-4x-9y-4=0$
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$x^2+y^2+4x+9y+4=0$
0%
$x^2+y^2+4x+6y+9=0$
0%
$x^2+y^2-4x-6y-9=0$

The normal at

$P(8, 8)$ to the parabola

$y^2=8x$ cuts it again at Q then PQ =

0%
$10$
0%
$10\sqrt5$
0%
$4\sqrt5$
0%
$50$

Circles are drawn passing through the origin

$O$ to intersect the coordinate axes at point

$P$ and

$Q$ such that

$m$ .

$PO+n.OQ=k$ , then the fixed point satisfy all of them, is given by

0%
$(m,n)$
0%
$\dfrac{m^{2}}{k},\dfrac{n^{2}}{k}$
0%
$\dfrac{mk}{m^{2}+n^{2}},\dfrac{nk}{m^{2}+n^{2}}$
0%
$(k,k)$

Explanation

${x}^{2}+{y}^{2}+2gx+2fy+c=0$

$C=0$

${x}^{2}+{y}^{2}+2gx+2fy=0$

${x}^{2}+2g\left(x\right)=0$

$x\left(x+2g\right)=0$

${x}^{2}+{y}^{2}+2gx+2fy=0\quad \quad H.O.P+nOQ=k$

For

$\left(m, n\right)\quad {m}^{2}+{n}^{2}+2gm+2fn\quad \quad H\left(-2g\right)+n\left(-2f\right)=k$

${m}^{2}+{n}^{2}-k\neq0\quad \quad 2mg-2n7=-k$

for

$\left(\dfrac{mk}{{m}^{2}+{n}^{2}}- \dfrac{nk}{{m}^{2}+{n}^{2}}\right)=$

$\left( \dfrac { { m }^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +k }^{ 2 } \right) }^{ 2 } } \quad \dfrac { n^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +n }^{ 2 } \right) }^{ 2 } } \right) +2g\left( \dfrac { mk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) +2f\left( \dfrac { nk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) \\$

$=\dfrac { { k }^{ 2 }\left( { m }^{ 2 }+{ n }^{ 2 } \right) +k\left( { m }^{ 2 }+{ n }^{ 2 } \right) \left( -k \right) }{ { \left( { m }^{ 2 }+{ n }^{ 2 } \right) }^{ 2 } }$

$\Rightarrow \left(\dfrac{mk}{{m}^{2}+{n}^{2}}, \dfrac{nk}{{m}^{2}+{n}^{2}}\right)$

1377871_1150152_ans_610a71c6bb86416ab263e5c8c9c075aa.png

The vertex

$A$ of the parabola

${y}^{2}=4ax$ is joined to any point

$P$ on it and

$PQ$ is drawn at right angles to

$AP$ to meet the axis in

$Q$ . Projection of

$PQ$ on the axis is equal to

0%
twice the length of latus rectum
0%
the latus length of rectum
0%
half the length of latus rectum
0%
one fourth of the length of latus rectum

If equation

$(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$ represents an ellipse, then

$\lambda \in$

0%
$(0, 1)$
0%
$(0, 2)$
0%
$(1, 2)$
0%
$(0, 1)\cup (1, 2)$

The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes

0%
$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=4$
0%
$\dfrac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$
0%
$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=4$
0%
none of these

If the curves

$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$ and

$\dfrac{x^{2}}{l^{2}}-\dfrac{y^{2}}{4}=1$ cut each other orthogonally then

$l^{2}=$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Equation of the latus rectum of the hyperbola

$(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$ is

0%
$y - 1/5 =-3/4(x - 1/2)$
0%
$x - 1/5 =-3/4(y - 1/2)$
0%
$y + 1/5 =-3/4(x + 1/2)$
0%
$x + 1/5 =-3/4(y + 1/2)$

If the equation

$\frac { \lambda ( x + 1 ) ^ { 2 } } { 3 } + \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$ represents a circle then

$\lambda =$

0%
1
0%
$\frac { 3 } { 4 }$
0%
0
0%
$- \frac { 3 } { 4 }$

The area enclosed between the parabolas

$y^2 = 4x$ and

$x^2 = 4y$ is

0%
$\dfrac{16}{3}$ sq. unit
0%
$\dfrac{3}{4}$ sq. unit
0%
$\dfrac{3}{16}$ sq. unit
0%
None of these

Equation of the curve passing through the point

$(1,\ 2)$ such that the intercept on the

$x-$ axis cut off between the tangent and origin is twice the abscissa is given by:

0%
$xy=2$
0%
$xy=1$
0%
$xy=2y$
0%
$xy=2x$

Let

$S$ and

$S ^ { 1 }$ are the foci of an ellipse whose eccentricity is

$\frac { 1 } { \sqrt { 2 } } , B$ and

$B ^ { 1 }$ are the ends of minor axis then

$S B S ^ { \prime } B ^ { 1 }$ forms

$a$ ________________.

0%
Parallelgram
0%
Rhombus
0%
Square
0%
Rectangle

$D.E$ of the curve for which the initial ordinates of any tangent is equal to the corresponding number

0%
Second degree in $x$
0%
Hemiohenous of second degree
0%
Has separable variables
0%
is of second degree

ABCD is a square of side 1 unit. A circle passes through vertices A,B of the square and the remaining two vertices of the square lie out side the circle. The length of the tangent draw to the circle from vertex D is 2 units. The radius of the circle is

0%
$\sqrt{5}$
0%
$\dfrac{1}{2}$ $\sqrt{10}$
0%
$\dfrac{1}{3}$ $\sqrt{12}$
0%
$\sqrt{8}$

The coordinates of the foci of the hyperbola

$xy=c^2$ are

0%
$(\pm C, \pm C)$
0%
$(\pm \dfrac{c}{\sqrt{2}}< \pm\dfrac{c}{\sqrt{2}})$
0%
$(\pm 2c, \pm 2c)$
0%
$(\pm \sqrt{2}c, \pm \sqrt{2}c)$

The locus of point of intersection

$P$ of tangents to ellipse

$2x^{2}+3y^{2}=6$ at

$A$ and

$B$ if

$AB$ subtend

$90^{o}$ angle at centre of ellipse is an ellipse whose eccentricity is equal to

0%
$\surd {5}/4$
0%
$\surd {5}/3$
0%
$2/\surd {5}$
0%
$none\ of\ these$

The equation of the circle, passing through the point

$\left(2,8\right)$ , touching the lines

$4x-3y-24=0$ and

$4x+3y-42=0$ and having x coordinate of the centre of the circle numerically less then or equal to 8 is

0%
$x^2+y^2+4x-6y-12=0$
0%
$x^2+y^2-4x+6y-12=0$
0%
$x^2+y^2-4x-6y-12=0$
0%
None of these

Explanation

Since

$C\left(\alpha,\beta\right)$ be the centre of the circle.

Since the circle passes through the point

$\left(2,8\right)$

$\therefore$ Radius of the circle

$=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$

Since the circle touches the lines

$4x-3y-24=0$ and

$4x+3y-42=0$

$\therefore\left|\dfrac{4\alpha-3\beta-24}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\left|\dfrac{4\alpha+3\beta-42}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$ ........

$\left(1\right)$

Solving them, we get

$4\alpha-3\beta-24=\pm\left(4\alpha+3\beta-42\right)$ ........

$\left(3\right)$

$\therefore 6\beta=18$ or

$\beta=3$ by taking the positive sign

and

$8\alpha=66$ or

$\alpha=\dfrac{66}{8}=\dfrac{33}{4}$ by taking negative sign.

Given

$\left|\alpha\right|\le 8\Rightarrow -8\le \alpha \le 8$

$\therefore \alpha \neq \dfrac{33}{4}$

Put

$\beta=3$ in equations

$\left(1\right)$ and

$\left(3\right)$ and equating, we get

${\left(4\alpha-33\right)}^{2}={\left(\alpha-2\right)}^{2}+25$

$\Rightarrow 16{\alpha}^{2}-264\alpha+1089=25{\alpha}^{2}+725-100\alpha$

$\Rightarrow 9{\alpha}^{2}+164\alpha-364=0$ which is quadratic in

$\alpha$

$\therefore \alpha=\dfrac{-164\pm\sqrt{{164}^{2}-36\times -364}}{2\times 9}=\dfrac{-164\pm\sqrt{{164}^{2}+36\times 364}}{18}$

$=\dfrac{-164\pm\sqrt{40000}}{18}=\dfrac{-164\pm 200}{18}=2,\dfrac{-182}{9}$

But

$-8\le \alpha\le 8$

$\therefore \alpha=2$

Now

${r}^{2}={\left(\alpha-2\right)}^{2}+{\left(3-8\right)}^{2}={\left(2-2\right)}^{2}+{\left(3-8\right)}^{2}=25$ where

$r$ is the radius of the circle

Hence the required equation of the required circle is

${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=25$

${x}^{2}+{y}^{2}-4x-6y+4+9-25=0$ or

${x}^{2}+{y}^{2}-4x-6y-12=0$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 10 - MCQExams.com

0:0:2

Practice Class 11 Engineering Maths Quiz Questions and Answers

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