MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 9

Find the equation of a circle whose centre is (2, - 1 ) an radius is 3

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

Explanation

Center

$=(2,-1)$ and

$r=3$ [ Given ]

We know that, center

$=(-g,-f)$

So, by comparing we get,

$g=-2$ and

$f=1$

$r=\sqrt{g^2+f^2-c}$

$\Rightarrow$

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

$\Rightarrow$

$9=4+1-c$

$\Rightarrow$

$9=5-c$

$\therefore$

$c=-4$

The equation of circle is

$\Rightarrow$

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

$\Rightarrow$

$x^2+y^2+2(-2)x+2(1)y-4=0$ [ Substituting values of

$g,f$ and

$c$ ]

$\Rightarrow$

$x^2+y^2-4x+2y-4=0$

The equation of a circle which is passing through the vertices of an equilateral triangle whose median is of length 3 a is

0%
$x ^ { 2 } + y ^ { 2 } = 9 a ^ { 2 }$
0%
$x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }$
0%
$x ^ { 2 } + y ^ { 2 } = 4 a ^ { 2 }$
0%
$x ^ { 2 } + y ^ { 2 } = a ^ { 2 }$

Explanation

Median of the equilateral triangle is

$3a$

$\Rightarrow=3a$

$\triangle ADB$ ,we have

$\Rightarrow(AB)^2=(AD)^2+(BD)^2$

$\Rightarrow(AB)^2=(3a)^2+\left(\dfrac{AB}{2}\right)^2$

$\Rightarrow(AB)^2=9a^2+\dfrac{(AB)^2}{4}$

$\Rightarrow\dfrac{3}{4}(AB)^2=9a^2$

$\Rightarrow(AB)^2=12a^2$

Now, In

$\triangle OBD,$

$\Rightarrow(OB)^2=(OD)^2+(BD)^2$

$\Rightarrow r^2=(3a-r)^2+\left(\dfrac{AB}{2}\right)^2$

$\Rightarrow r^2=9a^2+r^2-6ar+3a^2$

$[\because(AB)^2=10a^2]$

$\Rightarrow6ar=12a^2$

$\Rightarrow r=2a$

So, the equation of circle is

$\Rightarrow(x-0)^2+(y-0)^2=(2a^2)$

$\Rightarrow x^2+y^2=4a^2$

2119521_1459683_ans_48fc675a785247bf9dcb7e3f2e9a2da0.png

Equation

$\sqrt{(x-2)^2 + y^2} + \sqrt{(x+2)^2 + y^2}$ = 4 represents _____________.

0%
Parabola
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Ellipse
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Circle
0%
straight lines

$\left| z-{ z }_{ 0 } \right| =\dfrac { \left| a\overline { z } +a\overline { z } +b \right| }{ 2\left| a \right| }$ represent a parabola, Then the length of latus rectum of parabola

0%
$\left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right|$
0%
$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ \left| a \right| }$
0%
$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ 2\left| a \right| }$
0%
None of these

If the tangent to the curve,

$y=x^3+ax-b$ at the point

$(1, -5)$ is perpendicular to the line,

$-x+y+4=0$ , then which one of the following points lies on the curve?

0%
$(-2, 2)$
0%
$(2, -2)$
0%
$(2, -1)$
0%
$(-2, 1)$

Explanation

$y=x^2+ax-b$

$(1, -5)$ lies on the curve

$\Rightarrow -5=1+a-b\Rightarrow a-b=-6$ .(i)
Also,

$y'=3x^2+a$

$y'_{(1, -5)}=3+a$ (slope of tangent)

$\because$ this tangent is

$\perp$ to

$-x+y+4=0$

$\Rightarrow (3+a)(1)=-1$

$\Rightarrow a=-4$ .(ii)
By (i) and (ii):

$a=-4$ ,

$b=2$

$\therefore y=x^3-4x-2$

$(2, -2)$ lies on this curve.

In what ratio, the point of intersection of the common tangents to hyperbola

$\dfrac{x^2}{1}-\dfrac{y^2}{8}=1$ and parabola

$y^2=12x$ , divides the foci of the given hyperbola?

0%
$3:4$
0%
$3:2$
0%
$5:4$
0%
$5:3$

Explanation

Let equation of common tangent is

$y=mx+\dfrac{3}{m}$

$\therefore \left(\dfrac{3}{m}\right)^2=1.m^2-8$

$\Rightarrow m^4-8m^2-9=0$

$\Rightarrow m^2=9$

$\Rightarrow m=\pm 3$

$\therefore$ equation of common tangents are

$y=3x+1$ &

$y=-3x+1$

$\therefore \dfrac{PS}{PS'}=\dfrac{3+\dfrac{1}{3}}{-\dfrac{1}{3}+3}=\dfrac{5}{4}$ .
1246482_1613053_ans_3f816eb453834c11bf6d9983f1d1f33c.png

$AB$ is a chord of the circle

$x^{2} + y^{2} = 9$ . The tangent at

$A$ and

$B$ intersect at

$C$ . If

$(1, 2)$ is the midpoint of

$AB$ , the area of

$\triangle ABC$ is (in square units).

0%
$9$
0%
$\dfrac {7}{\sqrt {5}}$
0%
$\dfrac {9}{\sqrt {5}}$
0%
$\dfrac {8}{\sqrt {5}}$

Explanation

The equation of chord with midpoint

$(1, 2)$ in the circle

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

$x \times 1 + y\times 2 - 9 = 1^{2} + 2^{2} - 9$

$\Rightarrow x + 2y - 5 = 0 .... (1)$

Also, if the point of intersection of tangents

$C$ has coordinates

$(h, k)$ , then the equation of common chord is

$hx + ky - 9 = 0 .... (2)$

Since (1) and (2) represent the same equation, we get

$\dfrac {h}{1} = \dfrac {k}{2} = \dfrac {9}{5}$

$\Rightarrow h = \dfrac {9}{5}, k = \dfrac {18}{5}$

From figure,

$AD = \sqrt {OA^{2} - OD^{2}} = \sqrt {9 - 5} = 2$

$CD = OC - OD = \sqrt {h^{2} + k^{2}} - \sqrt {5} = \dfrac {9}{\sqrt {5}} - \sqrt {5} = \dfrac {4}{\sqrt {5}}$

The area of

$ABC = \dfrac {1}{2} AB . CD = AD.CD = 2\times \dfrac {4}{\sqrt {5}} = \dfrac {8}{\sqrt {5}}$ .

The equation

$\dfrac {x^{2}}{2 - \lambda} + \dfrac {y^{2}}{\lambda - 5} - 1 = 0$ represents an ellipse, if

0%
$\lambda < 5$
0%
$\lambda < 2$
0%
$2 < \lambda < 5$
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$\lambda < 2$ or $\lambda < 5$

Explanation

General equation of ellipse is

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

So both denominator should be positive as they are squares

In the given equation

$\dfrac {x^2}{2-\lambda} +\dfrac {y^2}{\lambda -5}-1=0$

So,

$2-\lambda > 0, +(\lambda -5) >o$

$\Rightarrow \ \lambda < 2, \lambda > 5$

$\Rightarrow \ 2 < \lambda < 5$

The eccentricity of the ellipse

$5x^{2} + 9y^{2} = 1$ is

0%
$\dfrac 23$
0%
$\dfrac 34$
0%
$\dfrac 45$
0%
$\dfrac 12$

Explanation

Consider the equation of the ellipse is

$5x^2+9y^2=1$

$\dfrac {\dfrac {x^2}{1}}{5}+\dfrac {\dfrac {y^2}{1}}{9}=1$

Comparing with the standard equation of the ellipse, we get:

$a^2=1/5$ and

$b^2=1/9$ i.e.,

$a=1/ \sqrt 5$ and

$b=1/3$

Here

$a > b$

now,

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

$\Rightarrow \ e=\sqrt {1-\dfrac {1/9}{1/5}}=\sqrt {1\times 5/9}=\sqrt {4/9}$

$e=2/3$

Hence eccentricity of ellipse

$=2/3$

The eccentricity of the ellipse

$9x^{2} + 25y^{2} - 18x - 100y - 116 = 0$ , is

0%
$25/16$
0%
$4/5$
0%
$16/25$
0%
$5/4$

Explanation

Given equation of ellipse is

$9x^2+25y^2-18x-100y-116=0$

$9(x^2-2x+1)+25(y^2-4y+4)-225=0$

$9(x-1)^2+25(y-2)^2=225$

$\dfrac{9(x-1)^2}{225}+\dfrac{25(y-2)^2}{225}=1$

$\Rightarrow \dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{9}=1$

comparing with the standard form

$\dfrac{(x-p)^2}{a^2}+\dfrac{(y-2)^2}{9}=1$

Here

$a^2>b^2$

eccentricity

$(e) =\sqrt{\dfrac{a^2-b^2}{a^2}}$

$e=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}}$

$e=\dfrac45$

An ellipse has its centre at

$(1, -1)$ and semi-major axis

$= 8$ and it passes through the point

$(1, 3)$ . The equation of the ellipse is

0%
$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
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$\dfrac {(x - 1)^{2}}{64} + \dfrac {(y + 1)^{2}}{16} = 1$
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$\dfrac {(x - 1)^{2}}{16} + \dfrac {(y + 1)^{2}}{64} = 1$
0%
$\dfrac {(x + 1)^{2}}{64} + \dfrac {(y - 1)^{2}}{16} = 1$

Explanation

Given that

centre is at

$(1, -1)$

semi major axis

$(a)=8$

so, equation of ellipse can be written as

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

It passes through point

$(1,3)$

i.e,

$x=1, y=3$

Putting these value in equation

$(1)$ we get

$\dfrac {(1-1)^2}{a^2} +\dfrac {(3+1)^2}{b^2}=1$

$\dfrac {16}{b^2}=1$

$b^2=16\ \Rightarrow b=4$

Substituting the values of

$a$ and

$b$ in equation

$(1)$ we get

$\dfrac {(x-1)^2}{64}+\dfrac {(y+1)^2}{16}=1$

This is the required equation of ellipse

The eccentricity of the ellipse

$25x^{2} + 16^{2} = 400$ is

0%
$\dfrac 35$
0%
$\dfrac 13$
0%
$\dfrac 25$
0%
$\dfrac 15$

Explanation

Given equation of ellipse is

$25x^2+16y^2=400$

$\dfrac{25x^2}{400}+\dfrac{16y^2}{400}=1$

$\Rightarrow \dfrac{x^2}{16}+\dfrac{y^2}{25}=1$

comparing with the standard form

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

Here

$b^2>a^2$

$\Rightarrow$ Eccentricity

$(e) =\sqrt{\dfrac{b^2-a^2}{b^2}}$

$e=\sqrt{\dfrac{25-16}{25}}=\sqrt{\dfrac{9}{25}}$

$e=\dfrac 35$

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

0%
$\cfrac{1}{\sqrt 3}$
0%
$\cfrac{1}{\sqrt 5}$
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$\cfrac{1}{\sqrt 2}$
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None of the above

If the major axis of an ellipse is three times the minor axis, then its eccentricity is equal to

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

Explanation

Given

major axis of an ellipse

$=3\times$ minor axis

$a=3b$

Eccentricity

$(e)=\sqrt{\dfrac{a^2-b^2}{a^2}}$

$e=\sqrt{\dfrac{(3b)^2-b^2}{(3b)^2}}=\sqrt{\dfrac{8b^2}{9b^2}}$

$e=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt 2}{3}$

Two eccentricity of ellipse is

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

If the latus-rectum of an ellipse is one half of its minor axis, then its eccentricity is

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{\sqrt {2}}$
0%
$\dfrac {\sqrt {3}}{2}$
0%
$\dfrac {\sqrt {3}}{4}$

Explanation

Consider the equation of the ellipse is

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

It is given that; length of latus rectum = half of minor axis

$\dfrac {2b^2}{a}=b \Rightarrow a=2b$

now,

$b^2=a^2 (1-e^2)$

$\Rightarrow \ b^2=4b^2 (1-e^2)\Rightarrow 1-e^2 =1/4 \Rightarrow e^2 =3/4$

$\therefore \$ eccentricity

$(e)=\sqrt 3 /2$

The eccentricity of the conic

$9x^{2} + 25y^{2} = 225$ is

0%
$\dfrac 25$
0%
$\dfrac 45$
0%
$\dfrac 13$
0%
$\dfrac 15$
0%
$\dfrac 35$

Explanation

Equation of the given ellipse is

$9x^2+25y^2=225$

Dividing through out by

$225$

$\dfrac {x^2}{25}+\dfrac {y^2}{9}=1$

Hence

$a^2=25$ and

$b^2=9$

$\therefore \ c=\sqrt {25-9}=\sqrt {16}=4$

eccentricity

$(e)=\dfrac {c}{a}=\dfrac {4}{5}$

The vertex of the parabola

$y^2 - 4y - x + 3 = 0$ is

0%
$(-1,3)$
0%
$(-1,2)$
0%
$(2,-1)$
0%
$(3, -1)$

Explanation

We have,

$y^2 - 4y - x + 3 = 0$

$\Rightarrow (y-2)^2 - 4 - x+3 = 0$

$\Rightarrow (y-2)^2 = (x+1)$

$\therefore$ Vertex of the parabola

$= (-1,2)$

The eccentricity of the conic

$9{x}^{2}-16{y}^{2}=144$ is

0%
$\cfrac{5}{4}$
0%
$\cfrac{4}{3}$
0%
$\cfrac{4}{5}$
0%
$\sqrt{7}$

The equation

$5x^2 + y^2 + y = 8$ represents

0%
An ellipse
0%
A parabola
0%
A hyperbola
0%
A Circle

Explanation

We have,

$5x^2 + y^2 + y = 8$

$\Rightarrow 5x^2 + \left(y + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2 = 8$

$\Rightarrow 5x^2 + \left(y + \dfrac{1}{2}\right)^2 = \dfrac{33}{8}$

$\Rightarrow \dfrac{\dfrac{5x^2}{33}}{8} + \dfrac{\dfrac{\left(y+\dfrac{1}{2}\right)^2}{33}}{8} = 1$

$\Rightarrow \dfrac{x^2}{\left(\dfrac{33}{40}\right)} + \dfrac{\left(y+\dfrac{1}{2}\right)^2}{\left(\dfrac{33}{8}\right)} = 1$
which is sn equation of ellipse.

Write the length of the latus-rectum of the hyperbola

$16{x}^{2}-9{y}^{2}=144$

0%
$\cfrac{5}{3}$
0%
$\cfrac{4}{3}$
0%
$\cfrac{3}{4}$
0%
$\cfrac{4}{5}$

The equation of the circle with centre

$(2, 2)$ which passes through

$(4,5)$ is

0%
$x^2 + y^2 - 4x + 4y - 77 = 0$
0%
$x^2 + y^2 - 4x - 4y - 5 = 0$
0%
$x^2 + y^2 + 2x + 2y - 59 = 0$
0%
$x^2 + y^2 - 2x - 2y - 23 = 0$
0%
$x^2 + y^2 + 4x - 2y -26 = 0$

Explanation

Radius of circle is

$\sqrt{(4-2)^2+(5-2)^2}=\sqrt{13}$ So, equation of circle is

$(x-2)^2+(y-2)^2=13$

$\Rightarrow x^2+4-4x+y^2+4-4y=13$

$\Rightarrow x^2+y^2-4x-4y-5=0$

The centre of the ellipse

$4x^2 + y^2 - 8x + 4y - 8 = 0$ is

0%
$(0,2)$
0%
$(2,-1)$
0%
$(2,1)$
0%
$(1,2)$

Explanation

We have,

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

$\Rightarrow (4x^2 - 8x) + (y^2 + 4y)-8=0$

$\Rightarrow (4x^2 - 2x) + (y^2 + 4y)-8=0$

$\Rightarrow 4[(x-1)^2 - 1]+[(y + 2)^2 - 4]-8=0$

$\Rightarrow 4(x-1)^2 -4 + (y + 2)^2 - 4 -8=0$

$\Rightarrow 4(x-1)^2 + (y+2)^2 = 16$

$\Rightarrow \dfrac{(x-1)^2}{4} + \dfrac{(y+2)^2}{16} = 1$

$\therefore$ Centre

$= (1,2)$

The latus-rectum of the hyperbola

$16{x}^{4}-9{y}^{2}=144$ is

0%
$16/3$
0%
$32/3$
0%
$8/3$
0%
$4/3$

The eccentricity of the ellipse

$\dfrac{(x-1)^2}{2} + \left(y + \dfrac{3}{4}\right)^2 = \dfrac{1}{16}$ is

0%
$\dfrac{1}{\sqrt{2}}$
0%
$\dfrac{1}{2\sqrt{2}}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

Explanation

$\dfrac{(x-1)^2}{2} + \left(y + \dfrac{3}{4}\right)^2 = \dfrac{1}{16}$

$\Rightarrow 8(x-1)^2 + 16\left(y +\dfrac{3}{4}\right)^2 = 1$

$\Rightarrow \dfrac{(x-1)^2}{\dfrac{1}{8}} + \dfrac{\left(y + \dfrac{3}{4}\right)^2}{\dfrac{1}{16}} = 1$

$\therefore a^2 = \dfrac{1}{8}$ and

$b^2 = \dfrac{1}{16}$

$\Rightarrow a = \dfrac{1}{2\sqrt{2}}$ and

$b = \dfrac{1}{4}$

$\therefore e = \sqrt{1 - \dfrac{b^2}{a^2}}$

$[\because a > b]$

$= \sqrt{1-\dfrac{\dfrac{1}{16}}{\dfrac{1}{8}}} = \sqrt{1 - \dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}$

The latus rectum of the parabola

$\displaystyle x = at^2 + bt + c, y = a't^2 + b't + c'$ is

0%
$\displaystyle \dfrac {\left ( aa'- bb' \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$
0%
$\displaystyle \dfrac {\left ( ab'- a'b \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$
0%
$\displaystyle \dfrac {\left ( bb'- aa' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$
0%
$\displaystyle \dfrac {\left ( a'b- ab' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$

The foci of the ellipse

$25\left ( x + 1 \right )^2 + 9\left ( y + 2 \right )^2 = 225$ , are at

0%
$\left ( -1 , 2 \right )$ and $\left ( -1 , -6 \right )$
0%
$\left ( -2 , 1 \right )$ and $\left ( -2 , 6 \right )$
0%
$\left ( -1 , -2 \right )$ and $\left ( -2 , -1 \right )$
0%
$\left ( -1 , -2 \right )$ and $\left ( -1 , -6 \right )$

The centre of a circle is

$C(2,-5)$ and the circle passes through the point

$A(3,2)$ . The equation of the circle is

0%
${ x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-21=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+4x-10y+21=0$
0%
none of these

Explanation

radius of the circle is

$AC=\sqrt { { (3-2) }^{ 2 }+{ (2+5) }^{ 2 } } =\sqrt { 1+49 } =\sqrt { 50 }$

equation of the circle is

${ (x-x_1) }^{ 2 }+{ (y-y_1) }^{ 2 }=r^2$

${ (x-2) }^{ 2 }+{ (y+5) }^{ 2 }=50\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$

If the parabola

${y}^{2}=4ax$ passes through the point

$P(3,2)$ , then the length of its latus rectum is

0%
$\cfrac{1}{3}$
0%
$\cfrac{2}{3}$
0%
$\cfrac{4}{3}$
0%
$4$

Explanation

Since the point

$P(3,2)$ lies on

${ y }^{ 2 }=4ax$ , we have

$4a\times 3={ 2 }^{ 2 }\Rightarrow a=\cfrac { 1 }{ 3 }$

Latus rectum

$=4a=4\times \cfrac { 1 }{ 3 } =\cfrac { 4 }{ 3 }$

The equation

$\displaystyle ax^2 + 4xy + y^2 + ax + 3y + 2 = 0$ represents a parabola if a is

0%
$\displaystyle -4$
0%
$\displaystyle 4$
0%
$\displaystyle 0$
0%
$\displaystyle 8$

The latus rectum of the hyperbola

$9x^{2} - 16y^{2} - 18x - 32y - 151 = 0$ is

0%
$\dfrac{9}{4}$
0%
9
0%
$\dfrac{3}{2}$
0%
$\dfrac{9}{2}$

Explanation

Hyperbola

$9x^{2} - 16y^{2} - 18x - 32y - 151 = 0$ can be written as

$9(x^{2}-2x) - 16 (y^{2}+2y) = 151$

$\Rightarrow 9(x-1)^{2} - 16(y+1)^{2} = 151 + 9 - 16 = 144$

$\Rightarrow \dfrac{(x-1)^{2}}{16} - \dfrac{(y+1)^{2}}{9} = 1$

$\dfrac{X^{2}}{16} - \dfrac{Y^{2}}{9} = 1$

$\left [ where X = x-1, Y= y+1 \right ]$

Here

$a^{2} = 16, b^{2} = 9$

Latus rectum

$= 2\dfrac{b^{2}}{a} = \dfrac{2(9)}{4} = \dfrac{9}{2}$

If the eccentricity of the ellipse

$\dfrac{x^{2}}{a^{2}+ 1}+ \dfrac{y^{2}}{a^{2}+ 2}=1 \, is \, \dfrac{1}{\sqrt{6}}$ , then latus rectrum of ellipse is

0%
$\dfrac{5}{\sqrt{6}}$
0%
$\dfrac{10}{\sqrt{6}}$
0%
$\dfrac{8}{\sqrt{6}}$
0%
none of these

Explanation

Here

$a^{2}+ 2 > a^{2}+ 1$

$a^{2}+ 1= (a^{2}+ 2)(1- e^{2})$

$a^{2}+1= (a^{2}+ 2)\dfrac{5}{6}$

$6a^{2}+ 6= 5a^{2}+ 10$

$a^{2}= 10-6=4$

$a=2$

Latus rectum

$=\dfrac{2(a^{2}+1)}{\sqrt{a^{2}+2}}= \dfrac{2 \times5}{\sqrt{6}}=\dfrac{10}{\sqrt{6}}$

The co- ordinates (2,3) and (1,5)are the foci of an ellipse which passes through the origin , then the equation of

0%
tangent at the origin is $(3 \sqrt{2}-5)x+ (1-2 \sqrt{2})y=0$
0%
tangent at the origin is $(3 \sqrt{2}+ 5)x+ (1+2 \sqrt{2}y)=0$
0%
normal at the origin is ( $3\sqrt{2}+ 5)x -(2 \sqrt{2}+1 )y=0$
0%
normal at the origin is x $(3 \sqrt{2}-5) -y(1- 2 \sqrt{2})=0$

Explanation

Tangent and normal are bisectors of

$\angle SPS'$

Now equation of SP is

$y= \dfrac{3x}{2}$ and that of S"P is y=5x

Then their equations are

$\dfrac{3x-2y}{\sqrt{13}}= \pm \dfrac{5x-y}{\sqrt{26}}$

$3x- 2y= \pm \dfrac{5x-y}{\sqrt{2}}$

lines are

$(3 \sqrt{2}- 5)x+ (1- 2\sqrt{2} )y=0\, and\, (3\sqrt{2}+ 5)x- (2\sqrt{2}+1 )y=0$

Now (2,3) and (1,5) lie on the same side of

$(3 \sqrt{2}- 5)x+ (1- 2 \sqrt{2})y=0$ , which is equation of tangent .

Points (2,3) and (1,5) lie on the different sides of

$(3\sqrt{2}+ 5)x- (2 \sqrt{2}+ 1 )y=0$ , which is equation of normal.

If the equation of the ellipse is

$3x^{2}+ 2 y^{2}+6x-8y+5=0,$ then which of the following is/ are true?

0%
$e=\dfrac{1}{\sqrt{3}}$
0%
center is (-1,2)
0%
foci are (-1,1) and (-1,3)
0%
directrices are $y= 2 \pm \sqrt{3}$

Explanation

$3x^{2}+ 2y^{2}+ 6x- 8y+ 5=0$

$\dfrac{(x+1)^{2}}{2}+ \dfrac{(y-2)^{2}}{3}=1$

Therefore, center is (-1,2) and ellipse is vertical (b>a)

$a^{2}=2, b^{2}=3$

Now

$2=3(1-e^{2})$

$e=\dfrac{1}{\sqrt{3}}$

Foci are(-1,2

$\pm \, be)\, and \, (-1,2 \pm 1)= (-1,3)\, and\, (-1,1)$ and directrix are

$y=2 \pm \dfrac{b}{e} \implies y=5\, and \, y=-1$

Consider the parabola whose focus is at (0,0) and tangent at vertex is

$x-y+1=0$

The length of latus rectum is

0%
$4 \sqrt{2}$
0%
$2 \sqrt{2}$
0%
$8 \sqrt{2}$
0%
$3 \sqrt{2}$

Explanation

The distance between the focus and the tangent at the

$\text { vertex }=\dfrac{|0-0+1|}{\sqrt{1^{2}+1^{2}}}=\dfrac{1}{\sqrt{2}}$

The directrix is the line parallel to the tangent at vertex and at a distance

$2 \times \dfrac{1}{\sqrt{2}}$ from the focus.

Let equation of directrix is

$x-y+\lambda=0$

$\text { where } \quad \dfrac{\lambda}{\sqrt{1^{2}+1^{2}}}=\dfrac{2}{\sqrt{2}}$

$\Rightarrow \quad \lambda=2$

Let

$P(x, y)$ be any moving point on the parabola, then

$O P =P M$

$\Rightarrow x^{2}+y^{2} =\left(\dfrac{x-y+2}{\sqrt{1^{2}+1^{2}}}\right)^{2}$

$\Rightarrow 2 x^{2}+2 y^{2}=(x-y+2)^{2}$

$\Rightarrow x^{2}+y^{2}+2 x y-4 x+4 y-4=0$

Latus rectum length

$=2 \times$ (distance of focus from directrix )

$=2\left|\dfrac{0-0+2}{\sqrt{1^{2}+1^{2}}}\right|$

$=2 \sqrt{2}$

Solving parabola with

$x$ -axis,

$x^{2}-4 x-4 =0$

$\Rightarrow x =\dfrac{4 \pm \sqrt{32}}{2}=2 \pm 2 \sqrt{2}$

$\Rightarrow$ Length of chord on

$x$ -axis is

$4 \sqrt{2}$

Since the chord

$3 x+2 y=0$ passes through the focus, it is focal chord.

Hence, tangents at the extremities of chord are perpendicular.

1665342_1769002_ans_dfdd476040d044c1aeb03ffce34ea0f6.png

The length of the latus rectum of the parabola whose focus is

$\left (\frac{u^{2}} {2g} \sin 2\alpha, -\frac{u^{2}} {2g} \cos 2 \alpha \right )$ and directrix is

$y = \frac{u^{2}} {2g}$ is

0%
$\frac{u^{2}} {2g} \cos^{2} \alpha$
0%
$\frac{u^{2}} {2g} \cos 2 \alpha$
0%
$\frac{2u^{2}} {2g} \cos 2 \alpha$
0%
$\frac{2u^{2}} {2g} \cos^{2} \alpha$

Explanation

Focus of parabola is

$\left (\frac{u^{2}} {2g} \sin 2\alpha, -\frac{u^{2}} {2g} \cos 2 \alpha \right )$ and directrix is

$y = \frac{u^{2}} {2g}$
Length of latus rectum

$= 2 \times$ distance of focus form directrix

$= 2 \times \left |\dfrac{-\frac{u^{2}} {2g} \cos 2 \alpha - \frac{u^{2}} {2g}} {\sqrt{1}} \right |$

$= \dfrac{2u^{2}} {g} \cos^{2} \alpha$

The centre of a circle whose end points of a diameter are

$(-6,3)$ and

$(6,4)$ is

0%
$(8,-1)$
0%
$(4,7)$
0%
$(0,\frac{7}{2})$
0%
$(4,\frac{7}{2})$

The graph of

$y=x^2$ is a straight line.

0%
True
0%
False

$\text{If (0, 0) be the vertex and } 3x- 4y+ 2 = 0 \text{ be the directrix of a parabola,}$

$\text{ then the length of its latus rectum is - }$

0%
$\dfrac{4}{5}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{8}{5}$
0%
$\dfrac{1}{5}$

Explanation

$\textbf{Find perpendicular distance of vertex from directrix of the parabola}$

$\text{Given vertex is } (0,0) \text{ and directrix is } 3x - 4y +2 = 0$

$\text{We have a = perpendicular distance of directrix of the parabola from vertex}$

$\therefore \text{ a} = \bigg|\dfrac{3(0)- 4(0) + 2}{\sqrt{3^2 +4^2}} \bigg|$

$\boldsymbol{\Bigg[\because \textbf{ Perpendicular distance of a point } (x_1, y_1) \textbf{ from a line } ax +by + c = 0 \textbf{ is } \bigg| \dfrac{ax_1 + by_1 + c}{\sqrt{a^2 +b^2}}\bigg| } \Bigg]$

$\Rightarrow \text{a} = \dfrac{2}{5}$

$\because \text{ Length of latus rectum = 4a}$

$\Rightarrow \text{Length of latus rectum } = 4 \cdot \dfrac{2}{5} = \dfrac{8}{5}$

$\textbf{Hence, Option 'C' is correct}$

The distance between the foci of an ellipse is 16 and eccentricity is

$\dfrac12$ . Length of the major axis of the ellipse is

0%
8
0%
64
0%
16
0%
32

$\text{The length of the latus rectum of the parabola } x= ay^2+by+c \text{ is -}$

0%
$\dfrac{a}{4}$
0%
$\dfrac{a}{3}$
0%
$\dfrac{1}{a}$
0%
$\dfrac{1}{4a}$

Explanation

$\textbf{Convert given equation of parabola into standard form}$

$\text{Given equation of parabola is } x=ay^2+by+c$

$\Rightarrow \dfrac{x}{a}=y^2+b\dfrac{y}{a}+\dfrac{c}{a}$

$\Rightarrow \bigg( y+\dfrac{b}{2a}\bigg)^2-\dfrac{b^2}{4a^2}+\dfrac{c}{a}=\dfrac{x}{a}$

$\Rightarrow \bigg( y+\dfrac{b}{2a}\bigg)^2=\dfrac{1}{a}\bigg(x+\dfrac{b^2}{4a}-c\bigg)$

$\text{For a parabola of the form } y^2 = 4ax$

$\because \text{ Length of latus rectum = 4a}$

$\text{ Hence Length of latus rectum of given parabola is} = \dfrac{1}{a}$

$\textbf{Hence, Option 'C' is correct}$

The hyperbola

$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ has its conjugate axis of length

$5$ and passes through the point

$(2, 1)$ . The length of latus rectum is :

0%
$\dfrac{5}{4}\sqrt{29}$
0%
$\dfrac{5}{8}\sqrt{29}$
0%
$\sqrt{29}$
0%
$\dfrac{\sqrt{29}}{4}$

Explanation

$b=\dfrac{5}{2}$

$(2,1)$ lies on

$\dfrac{x^{2}}{a^{2}}-\dfrac{4y^{2}}{25}=1$

$\therefore \dfrac{4}{a^{2}}=\dfrac{29}{25}$

$\Rightarrow a=\dfrac{10}{\sqrt{29}}$
Latus rectum

$\dfrac{2b^{2}}{a}=2\left ( \dfrac{25}{4} \right ).\dfrac{\sqrt{29}}{10}$

$=\dfrac{5}{4}\sqrt{29}$

The graph of the curve

$x^2 + y^2 - 2xy - 8x - 8y + 32 = 0$ falls wholly in the

0%
first quadrant
0%
second quadrant
0%
third quadrant
0%
none of these

Explanation

$x^2 + y^2 - 2xy - 8x - 8y + 32 = 0$

$\Rightarrow (x-y)^2 = 8 (x + y - 4)$
is a parabola whose axis is

$x - y = 0$ and the tangent at the vertex is

$x + y - 4 = 0$ .
Also, when

$y = 0$ , we have

$x^2 - 8x + 32 = 0$
which gives no real values of

$x$ .
when

$x = 0$ , we have

$y^2 - 8y + 32 = 0$ which gives no real values of

$y$ .
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.

$f(\displaystyle \mathrm{m}_{\mathrm{i}}, \frac{1}{\mathrm{m}_{\mathrm{i}}})$ ,

$\mathrm{i}=1,2,3,4$ are four distinct points on the circle with centre origin, then value of

$\mathrm{m}_{1}\mathrm{m}_{2}\mathrm{m}_{3}\mathrm{m}_{4}$ is equal to

0%
$0$
0%
$-1$
0%
$1$
0%
$-a^{2}$

Explanation

Equation of circle having centre of origin $(0,0)$ and radius $=r$ ,

$S_{1};x^{2}+y^{2}=r^{2}----(1)$

$\therefore f(m_{1},\dfrac{1}{m_{2}}),f(m_{2},\dfrac{1}{m_{2}}),--f(m_{4},\dfrac{1}{m_{4}})$
These points lie on $S_{1}$ .

Let $\displaystyle f(m, \frac{1}{m})$ is point lie on S_{1},

$m^{2}+\dfrac{1}{m^{2}}=r^{2}$

$m^{4}+1-r^{2}m^{2}=0$

$m^{4}-1-r^{2}m^{2}+1=0$

$m_{1},m_{2},m_{3}$ and $m_{4}$ are roots of this equation

So, $m_{1}m_{2}m_{3}m_{4} =1$

lf the equation

$136 (x^{2}+y^{2})=(5x+3y+7)^{2}$ represents a conic, then its length of latus rectum is

0%
$\displaystyle \frac{7}{2\sqrt{34}}$
0%
$\displaystyle \frac{7}{\sqrt{34}}$
0%
$\displaystyle \frac{14}{\sqrt{34}}$
0%
$\displaystyle \frac{9}{\sqrt{34}}$

Explanation

$136 (x^{2}+y^{2})=(5x+3y+7)^{2}$

$\Rightarrow \displaystyle (x-0)^2+(y-0)^2= \dfrac{1}{136}(5x+3y+7)^{2}$

$\Rightarrow \displaystyle (x-0)^2+(y-0)^2= \left( \frac{1}{2}\right)^2 \left(\frac{5x+3y+7}{\sqrt{34}} \right)^2$

From the above form, we get focus of the conic has coordinates $(0,0)$ , and directrix has equation $5x+3y+7=0$ and eccentricity $e=\dfrac{1}{2}$

Distance between focus and directrix $=\dfrac{7}{\sqrt{34}}$

Length of latus rectum $=2\dfrac{b^2}{a}=2 \times e \times$ distance bet. focus and directrix $=2 \times \dfrac{1}{2} \times \dfrac{7}{\sqrt{34}}$

$= \dfrac{7}{\sqrt{34}}$

Hence, option B.

A point

$P(x, y)$ moves in

$XY$ plane such that

$x = a\cos^2 \theta$ and

$y = 2a \sin \theta$ , where

$\theta$ is a parameter. The locus of the point

$P$ is

0%
circle
0%
ellipse
0%
unbounded parabola
0%
part of a parabola

Explanation

$y^2=4a^{2}\sin^{2}{\theta}=$

$=4a^{2}-4ax=-4a(x-a)$
Now, since

$0\le\cos^{2}{\theta}\le{1}$

$\Rightarrow 0\le x \le a$
Also,

$-1\le\sin{\theta}\le{1}$

$\Rightarrow -2a\le y \le 2a$
So, the locus of

$P$ is a part of parabola.

Let P point on the circle

$x^2 + y^2 = 9$ , Q a point on the line

$7x + y + 3 = 0$ , and the perpendicular bisector of PQ be the line

$x - y + 1 = 0$ . Then the coordinate of P are

0%
(0, -3)
0%
(0, 3)
0%
$\displaystyle \left ( \frac{72}{25}, -\frac{21}{25} \right)$
0%
$\displaystyle \left ( -\frac{72}{25}, \frac{21}{25} \right)$

Explanation

Any point on the lines

$7x + y + 3 = 0$ is

$Q (t, -3 -7 t), t \in R.$
Now P

$(h, k)$ is image of point Q in the line

$x - y +1 = 0$
Then,

$\displaystyle \frac{h - t}{1} = \frac{k - (-3 - 7 t)}{-1}$

$= \displaystyle - \frac{2 (t - (-3 - 7t) + 1)}{1+1}$

$= -8t - 4$

$\Rightarrow (h, k) \equiv (-7t - 4, t + 1)$
This point lies on the circle

$x^2 + y^2 = 9$

$\Rightarrow (-7t - 4)^2 + (t + 1)^2 = 9$

$\Rightarrow 50 r^2 + 58t + 8 = 0$

$\Rightarrow 25r^2 + 19t + 4 = 0$

$\Rightarrow (25 t + 4) (t + 1) = 0$

$\Rightarrow t = -4/25, t = 1$

$\Rightarrow (h, k) = \displaystyle \left ( -\frac{72}{25}, \frac{21}{25} \right )$ or

$(3, 0)$

The length of latus rectum of the parabola whose parametric equations are

$x = t^{2} + t + 1$ ,

$y = t^{2}- t + 1$ , where

$t \in R$ , is equal to?

0%
$\sqrt{2}$
0%
$\sqrt{4}$
0%
$\sqrt{8}$
0%
$\sqrt{6}$

Explanation

$x=t^2+t+1$

$y=t^2-t+1$

Adding above equations, we get

$\therefore t=\dfrac{x-y}{2}$

Substituting this in first equation, we get

$\therefore x=\left(\dfrac{x-y}{2}\right)^2+\dfrac{x-y}{2}+1$

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

$\therefore (x-y)^2=2(x+y-2)$

So, length of latus rectum = 2

For the variable, the locus of the point of intersection of the lines

$3tx-2y+6t=0$ and

$3x+2ty-6=0$ is

0%
the ellipse $\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 9 } =1$
0%
the ellipse $\cfrac { { x }^{ 2 } }{ 9 } +\cfrac { { y }^{ 2 } }{ 4 } =1$
0%
the hyperbola $\cfrac { { x }^{ 2 } }{ 4 } -\cfrac { { y }^{ 2 } }{ 9 } =1$
0%
the hyperbola $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 4 } =1$

Explanation

Given equation of lines are

$3tx-2y+6t=0......(i)$
and

$3x+2ty-6=0.....(ii)$

On multiplying Eq

$(i)$ by

$t$ and then adding in eq.

$(ii)$ , we get

$(3{t}^{2}+3)x+6{t}^{2}-6=0$

$\Rightarrow$

$x=\cfrac{2(-1{t}^{2})}{(1+{t}^{2})}$

$\Rightarrow$

$x+x{t}^{2}=2-2{t}^{2}$

$\Rightarrow$

$(x+2){t}^{2}=(2-x)$

$\Rightarrow$

${t}^{2}=\cfrac{2-x}{2+x}.....(iii)$

On multiplying eq

$(ii)$ by

$t$ and then subtract from eq

$(i)$ we get

$(-2-2{t}^{2})y+6t+6t=0$

$12t=2(1+{t}^{2})y$

On squaring both sides, we get

$144{t}^{2}=4{y}^{2}{(1+{t}^{2})}^{2}$

$\Rightarrow$

$144\left(\cfrac{2-x}{2+x}\right)=4{y}^{2}{\left(1+\cfrac{2-x}{2+x}\right)}^{2}$ (from Eq

$(iii)$ ]

$\Rightarrow$

$36\left(\cfrac{2-x}{2+x}\right)={y}^{2}{\left(\cfrac{4}{2+x}\right)}^{2}$

$\Rightarrow$

$36\cfrac{2-x}{2+x}=\cfrac{16{y}^{2}}{{(2+x)}^{2}}$

$\Rightarrow$

$36(4-{x}^{2})=16{y}^{2}$

$\Rightarrow$

$9(4-{x}^{2})=4{y}^{2}$

$\Rightarrow$

$36-9{x}^{2}=4{y}^{2}$

$\Rightarrow$

$9{x}^{2}+4{y}^{2}=36$

$\Rightarrow$

$\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 9 } =1$ which represents an ellipse

If the line

$3x+4y=24$ and

$4x+3y=24$ intersects the coordinates axes at

$A,B,C$ and

$D$ , then the equation of the circle passing through these

$4$ points is

0%
$x^{2}+y^{2}-12x-12y+48=0$
0%
$x^{2}+y^{2}-14x-14y+48=0$
0%
$x^{2}+y^{2}-12x-12y+46=0$
0%
$x^{2}+y^{2}-14x-14y+46=0$

Explanation

writing in intercept form, we get

$\dfrac{x}{8}+\dfrac{y}{6}=1$ and

$\dfrac{x}{6}+\dfrac{y}{8}=1$
Hence the point are

$(0,8),(0,6)$ and

$(6,0)(8,0)$
Hence
The circle cuts the x axis at

$(6,0)(8,0)$
Hence x-coordinate of the center will be

$=6+\dfrac{8-6}{2}$

$=7$
The circle cuts the x axis at

$(0,6)(0,8)$
Hence y-coordinate of the center will be

$=6+\dfrac{8-6}{2}$

$=7$
The center of the circle will lie at

$(7,7)=C$
Now let the point

$(6,0)$ be A.
Hence
R=radius

$CA$

$=\sqrt{50}$
Hence, the equation of the circle will be

$(x-7)^{2}+(y-7)^{2}=50$

$x^{2}+y^{2}-14x-14y+98=50$
Or

$x^{2}+y^{2}-14x-14y+48=0$

The equation of a straight line drawn through the focus of the parabola

$y^2=-4x$ at an angle of

$120^o$ to the

$x$ -axis is.

0%
$y+\sqrt 3(x-1)=0$
0%
$y-\sqrt 3(x-1)=0$
0%
$y+\sqrt 3(x+1)=0$
0%
$y-\sqrt 3(x+1)=0$

Explanation

Parabola,

${ y }^{ 2 }=-4x$
Comparing above equation with

${ y }^{ 2 }=-4ax$

$4a=4\Rightarrow a=1$
Focus of

${ y }^{ 2 }=-4ax$ is

$\left( -a,0 \right)$

$\therefore$ Focus is

$\left( -1,0 \right)$
Now, line makes an angle of

${ 120 }^{ 0 }$ to the

$x$ - axis
Therefore,

$m=\tan { \theta } =\tan { { 120 }^{ 0 } }$

$\Rightarrow m=\tan { \left( { 180 }^{ 0 }-{ 60 }^{ 0 } \right) } =-\tan { { 60 }^{ 0 } }$

$\Rightarrow m=-\sqrt { 3 }$
Equation of line passing through

$\left( -1,0 \right)$ having slope

$-\sqrt { 3 }$ is

$:$

$y-0=-\sqrt { 3 } \left( x+1 \right)$

$\Rightarrow \boxed { y+\sqrt { 3 } \left( x+1 \right) =0 }$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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