MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 6

If a be the radius of a circle which touches x-axis at the origin, then its equation is

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${ x }^{ 2 }+{ y }^{ 2 }+ax=0$
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${ x }^{ 2 }+{ y }^{ 2 }\pm 2ya=0$
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${ x }^{ 2 }+{ y }^{ 2 }\pm 2xa=0$
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${ x }^{ 2 }+{ y }^{ 2 }+ya=0$

Explanation

The equation of the circle with centre at

$\left(h,k\right)$ and radius equal to

$a$ is

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

When the circle passes through the origin and centre lies on

$x-$ axis

$\Rightarrow h=a$ and

$k=0$

Then the equation

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

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

If a circle passes through the origin and centre lies on

$x-$ axis then the abscissa will be equal to the radius of the circle and the

$y-$ co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form

${\left(x\pm a\right)}^{2}+{y}^{2}={a}^{2}$

$\Rightarrow {x}^{2}+{a}^{2}\pm 2ax+{y}^{2}={a}^{2}={x}^{2}+{y}^{2}\pm 2ax=0$ is the required equation of the circle.

The centre of the circle

$(x-a)(x-b)+(y-c)(y-c)=0$ is

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$(a+b,c+d)$
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$9-a-b,-c-d)$
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$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$
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none of these

Explanation

Given the equation of the circle is

$(x-a)(x-b)+(y-c)(y-c)=0$

or,

$x^2-x(a+b)+ab+y^2-x(c+c)+cd=0$

or,

$x^2-2.x.\dfrac{a+b}{2}+\dfrac{(a+b)^2}{4}+y^2-2.y.\dfrac{c+d}{2}+\dfrac{(c+d)^2}{4}=\dfrac{(a+b)^2}{4}-ab+\dfrac{(c+d)^2}{4}-cd$

or,

$\left(x-\dfrac{a+b}{2}\right)^2+\left(y-\dfrac{c+d}{2}\right)^2=\dfrac{(a-b)^2}{4}+\dfrac{(c-d)^2}{4}$ .

From this equation it is clear that the centre is

$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$ .

Cantres of the three circles

$x^{2}+y^{2}-4x-6y-14=0$

$x^{2}+y^{2}+2x+4y-5=0$
and

$x^{2}+y^{2}-10x-16y+7=0$

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Are the vertices of a right triangle
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The vertices of an isosceles triangle which is not regular
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Vertices of a regular triangle
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Are collinear

Explanation

Centre of

$x^2+y^2-4x-6y-14=0\space is\space (-2,-3)$

Centre of

$x^2+y^2+2x+4y-5=0\space is\space (1,2)$

Centre of

$x^2+y^2-10x-16y+7=0\space is\space (-5,-8)$

Let these centres be

$A(-2,-3);B(1,2);C=(-5,-8)$

Now we find the lengths of

$AB, BC, CA$

$\Rightarrow AB=\sqrt{(-2-1)^2+(-3-2)^2}=\sqrt {34}$

$\Rightarrow BC=\sqrt{(-5-1)^2+(-8-2)^2}=\sqrt {136}$

$\Rightarrow CA=\sqrt{(-2+5)^2+(-3+8)^2}=\sqrt {34}$

Here we can see that

$AB+CA=BC$

Therefore

$ABC$ is a straight line.

Equation of the circle with centre on the

$y-$ axis and passing through the origin and the point

$(2,3)$ is

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$x^{2}+y^{2}+13y=0$
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$3x^{2}+3y^{2}+13x+3=0$
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$6x^{2}+6y^{2}-13y=0$
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$x^{2}+y^{2}+13+3=0$

Explanation

We have to find equation of a circle with center on the y-axis.

General equation of such circle is

$(x-0)^{2}+(y-k)^{2}=k^{2}$

It passes through

$(2,3)$

i.e,

$2^{2}+(3-k)^{2}=k^{2}$

$\Rightarrow 4+9+k^{2}-6k=k^{2}$

$\Rightarrow k=\dfrac{13}{6}$

$\therefore$ Equation of circle

$=x^{2}+\left(y-\dfrac{13}{6}\right)^{2}=\left(\dfrac{13}{6}\right)^{2}$

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

If the vertex

$= (2, 0)$ and the extremities of the latus rectum are

$(3, 2)$ and

$(3, -2)$ , then the equation of the parabola is

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$y^2 = 2x - 4$
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$x^2 = 4y - 8$
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$y^2 = 4x - 8$
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None

Explanation

$y^2=4ax$

$(y-0)^2=4a(x-2)$

$y^2=4a(x-2)$

$y^2=4(x-2)$

$y^2=4x-8$

The graph of curve

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

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first quadrant
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second quadrant
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third quadrant
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none of these

Explanation

${x}^{2}+{y}^{2}−2xy−8x−8y+32=0$

$\Rightarrow\,{x}^{2}+{y}^{2}−2xy=8x+8y-32$

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

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.

The eccentricity of the hyperbola whose latus-return is

$8$ and length of the conjugate axis is equal to half the distance between the foci, is

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$\dfrac43$
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$\dfrac4{\surd 3}$
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$\dfrac2{\surd 3}$
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$None\ of\ these$

Explanation

Given that the length of the latus rectum is

$8$ and length of the conjugate axis is equal to half the distance between the foci.

$\Rightarrow \dfrac{2b^2}{a}=8$ and

$2b=\dfrac{1}{2}(2ae)$

$\therefore \dfrac{2}{a}\left(\dfrac{ae}{2}\right)^2=8$

$\Rightarrow ae^2=16$ ...(1)

We have

$\dfrac{2b^2}{a}=8$

$\Rightarrow b^2=4a$

$\Rightarrow a^2(e^2-1)=4a$

$\Rightarrow ae^2-a=4$

$\Rightarrow 16-a=4$ (by (1))

$\Rightarrow a=12$

Substitute

$a=12$ in (1)

$\Rightarrow 12e^2=16$

$\Rightarrow e^2=\dfrac{4}{3}$

$\therefore e=\dfrac{2}{\sqrt{3}}$

$(4,3)$ and

$(-12,-1)$ are end points of a diameter of a circle, then the equation of the circle is-

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${ x }^{ 2 }+{ y }^{ 2 }-8x-2y-51=0$
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${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$
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${ x }^{ 2 }+{ y }^{ 2 }+8x+2y-51=0$
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None of these

Explanation

End points of diameter are

$(4,3)$ and

$(-12,-1)$

$\therefore$ Center of circle

$=\left( \cfrac { 4-12 }{ 2 } ,\cfrac { 3-1 }{ 2 } \right)$

$=(-4,1)$

Radius

$=\cfrac { 1 }{ 2 } \sqrt { { \left( 4+12 \right) }^{ 2 }+{ \left( 3+1 \right) }^{ 2 } }$

$=\sqrt { 68 }$

$\therefore$ Equation of circle is

${ \left( x+4 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( \sqrt { 68 } \right) }^{ 2 }$

${ x }^{ 2 }+8x+16+{ y }^{ 2 }-2y+1=68$

${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$

if the lines

$3x-4y-7=0$ and

$2x-3y-5=0$ are two diameter of a circle of area

$49\pi$ square units the equation of the circle is

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$x^2+y^2+2x-2y-62=0$
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$x^2+y^2-2x+2y-62=0$
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$x^2+y^2-2x+2y-47=0$
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$x^2+y^2+2x-2y-47=0$

Explanation

Given that,

Equations of diameter $3x-4y-7=0$ …… (1) and $2x-3y-5=0$ …… (2)

Solve equations are

$3x-4y-7=0$ $×2$

$2x-3y-5=0$ $×3$

$6x-8y-14=0$

$\underline{6x-9y-15=0}\,\,\,\,\,\,\,\,on\,\,subtracting\,\,\,that$

$y+1=0$

$y=-1$

Put the value of in equation (1)

$3x-4y-7=0$

$3x-4\left( -1 \right)-7=0$

$3x+4-7=0$

$3x-3=0$

$x=1$

Hence , the coordinates of the centre is $(1,-1)$

Also given area of circle= $49\pi$

$\pi {{r}^{2}}=49\pi$

${{r}^{2}}=49$

$r=7$

Then, we know that the equation of circle is

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

${{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{7}^{2}}$

${{x}^{2}}+1-2x+{{y}^{2}}+1+2y=49$

${{x}^{2}}+{{y}^{2}}-2x+2y+2=49$

${{x}^{2}}+{{y}^{2}}-2x+2y=49-2$

${{x}^{2}}+{{y}^{2}}-2x+2y=47$

${{x}^{2}}+{{y}^{2}}-2x+2y-47=0$

Hence, it is complete solution.

Option $(C)$ is correct answer.

Equation of the ellipse whose axes are the axes of coordinates and which passes through the point

$(-3,1)$ and has eccentricity

$\sqrt {\frac{2}{5}}$ is

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$5x^3+3y^2-48=0$
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$3x^2+5y^2-15=0$
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$5x^2+3y^2-32=0$
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$3x^2+5y^2-32=0$

Explanation

We know that equation of ellipse is

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

Given that

$e=\sqrt{\dfrac{2}{5}}$

$\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\dfrac{2}{5}}$

Taking square both side and solving , we get

$5{{a}^{2}}-5{{b}^{2}}=2{{a}^{2}}$

${{a}^{2}}=\dfrac{5{{b}^{2}}}{3}$ …….(2)

$\because$ ellipse pass through (-3,1)

Then $x=-3, y=1$

Put in equation (1) we get

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

${{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}}$

$\dfrac{5{{b}^{2}}}{3}+9{{b}^{2}}=\dfrac{5{{b}^{2}}}{3}.{{b}^{2}}$

${{b}^{2}}=\dfrac{32}{5}$ (From equation (1) and (2) )

Put in equation (2) , we get ${{a}^{2}}=\dfrac{32}{3}$

the value of a and b put in equation (1), we get

$\dfrac{{{x}^{2}}}{\dfrac{32}{3}}+\dfrac{{{y}^{2}}}{\dfrac{32}{5}}=1$

$3{{x}^{2}}+5{{y}^{2}}=32$

This is required equation

The name of the conic represented by

$\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=1$ is

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Circle
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Parabola
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Ellipse
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Hyperbola

Explanation

$\sqrt{x/a}+\sqrt{y/b}=1$

rearranging :

$\sqrt {y/b}=1-\sqrt{x/a}$

squaring

$y/b=1+x/a-2\sqrt{x/a}$

$\Rightarrow 2\sqrt{x/a}=1+x/a-y/b$

squaring again

$4x/a=1+x^2/a^2+y^2/b^2+2x/a-\dfrac{2xy}{ab}-\dfrac{2y}{b}$

Generally, when there is an equation of the form

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ , if

$B^2-4AC=0$ , then the centre is one of the following:

(1) A parabola

(2)

$2$ parallel lines ( certainly this is not the case)

(3)

$1$ line ( certainly this is not the case)

(4) No curve ( again not the case).

$\therefore \sqrt{x/a}+\sqrt{y/b}=1$ is a part of a parabola

The ellipse

$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ cuts x axis at A and y axis at B and the line joining the focus S and B makes an angle

$\dfrac{{3\pi }}{4}$ with x-axis. Then the eccentricity of the ellipse is

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

Explanation

REF.Image

B is (0,b) S is (ae,0)

$\dfrac{-b}{ae}=tan\dfrac{3\pi }{4}$

$e=\dfrac{b}{a}$

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

$\dfrac{b}{a}=\dfrac{1}{\sqrt{2}}$

$e=\dfrac{1}{\sqrt{2}}$

A is correct

1112369_1072341_ans_283dd6bffb1c46929ae4fa4b010c3928.jpg

The equation of the tangent to the ellipse such that sum of perpendiculars dropped from foci is 2 units, is

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$y cos3\pi/ 4 - x sin 3\pi /4=1$
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$y sin \frac{3\pi}{8}- x cos \frac{3\pi}{8}=1$
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$x cos \pi /8 - sin \pi /8=1$
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$y cos \frac{5\pi}{8}+x sin \frac{5\pi}{8}=1$

$(a, c)$ and

$(b, c)$ are the centres of two circles whose radical axis is the y-axis. If the radius of first circle is

$r$ then the diameter of the other circle is

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$2\sqrt {{r^2} - {b^2} + {a^2}}$
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$\sqrt {{r^2} - {a^2} + {b^2}}$
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$\left( {{r^2} - {b^2} + {a^2}} \right)$
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$2\sqrt {{r^2} - {a^2} + {b^2}}$

Explanation

Let radius of after circle be

$R$

$C_1:(x-a)^2+(y-c)^2=r^2$

$C_2:(x-b)^2+(y-c)^2=R^2$

Radius axis

$\to x=0$ OR

$c_1-c_2=0$

$c_1-c_2=x$

$2bx-2ax-r^2+R^2+a^2-b^2=x$

Put

$x=0$

$R^2=r^2+b^2-a^2$

$R=\sqrt {r^2+b^2-a^2}$

Diameter

$=2\sqrt {r^2 +b^2 -a^2}$

$D$ is correct

The equation of the latus rectum of the hyperbola

$\dfrac{(x-4)^2}{16}-\dfrac{(y-3)^2}{20}=1$ are?

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$x=1\pm 5$
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$x=4\pm 6$
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$y=2\pm 6$
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$y=3\pm 5$

Explanation

We are given the hyperbola

$\dfrac{x-4)^2}{16} - \dfrac{(y-3)^2}{20} = 1$ having the Focus at

$(\pm ae, 3)$

Where

$a = 4 \, \& \, b = 2\sqrt{5}$

$e = \dfrac{\sqrt{a^2+ b^2}}{a} = \dfrac{\sqrt{16 + 20}}{4} = \dfrac{3}{2}$

Sp, the lats rectum is the line through the focus and parallel to the directer

$x - x_1 = \pm ae$

here

$x_1$ is nothing but

$4$ .

$x = x_1 \pm ae = 4 \pm 4. \left(\dfrac{3}{2}\right) = 4 \pm 6$

$x = 4 \pm 6$

1282264_1085973_ans_fce26dc8a6df46eea8bc95e48c4aa744.png

The locus of the mid point of the focal radii of a variable point moving on the parabola,

${y^2}={8x}$ is a parabola whose

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Latus rectum is half the latus rectum of the original parabola
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Vertex is $(1,0)$
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Directrix is $y-axis$
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Focus has the co-ordinates $(2,0)$

Explanation

Let any point on parabola be

$(2{t}^{2},4t)$

focus is

$(2,0)$

midpoint is

$(1+{t}^{2},2t)$

Locus of midpoint is

${y}^{2}=4(x-1)$

Vertex is

$(1,0)$

L.R

$=4$

Diretrix is

$x-1=-1$

$\Rightarrow$

$x=0$ (y-axis)

Focus is

$(1,0)$

L.R original parabola

$=4\times 2=8$

Equation of circle having centre

$(5, 2)$ and which passes through the point

$(1, -1)$ is?

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$x^2+y^2-10x-4y-4=0$
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$x^2+y^2+10x+4y+4=0$
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$x^2+y^2-10x-4y-2=0$
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$x^2+y^2-10x-4y+4=0$

${x^2} - {y^2} + 5x + 8y - 4 = 0$

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rectangular hyperbola
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ellipse
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hyperbola
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pair of lines

S and S' foci of an ellipse. B is one end of the minor axis. If

$\angle{SBS'}$ is a right angled isosceles triangle, then e

$=?$

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

Explanation

We have

$S=(ae,0)\quad S'(-ae,0)and B=(0,b)$

Since it is given that

$\angle SBS'=90^o$

Slope of SB

$\times$ Slope of S'B

$=-1$

$\left(\dfrac{b-0}{0-ae}\right)\times\left(\dfrac{b-0}{b+ae}\right)=-1$

$\left(\dfrac{-b}{ae}\right)\left(\dfrac{b}{ae}\right)=-1$

$b^2=a^2e^2$

But,

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

So,

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

$1-e^2=e^2$

$2e^2=1$

$e^2=\dfrac{1}{2}$

$e=\dfrac{1}{\sqrt2}$

The latusrectum of a parabola

$y^{2}=4ax$ whose focal chord is

$PSQ$ such that

$SP=3$ and

$SQ=2$ , is given by

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$\dfrac{24}{5}$
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$\dfrac{12}{5}$
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$\dfrac{6}{5}$
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$\dfrac{1}{5}$

Explanation

Given that:

$SP=3,SQ=2$

Say the latus rectum is

$c$ then

$SP,\dfrac{c}{2},SQ$ are in

$H.P.$

$\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{c}$

$\dfrac{2+3}{6}=\dfrac{4}{c}$

$c=\dfrac{24}{5}.$

Hence the latus rectum is

$\dfrac{24}{5}.$

If the focus of a parabola divided a focal chord of the parabola in segment of length

$3$ and

$2$ the length of the latus rectum of the parabola is ?

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$\dfrac{3}{2}$
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$\dfrac{6}{5}$
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$\dfrac{12}{5}$
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$\dfrac{24}{4}$

Explanation

Let

$s(a,o)$ be the focus of the given parabola

Let the end points of the focal chord be

$P\left( a{ t }^{ 2 },2at \right) \quad and\quad Q\left( \dfrac { a }{ { t }^{ 2 } } ,\dfrac { -2a }{ t } \right)$

$SP$ and

$SQ$ are segments of the focal chord with length

$b$ and

$c$ respectively

therefore,

$b=3$ and

$c=2$

also,

$SP=\sqrt { { \left( a-at \right) }^{ 2 }+4{ a }^{ 2 }{ t }^{ 2 } } =\quad a\left( 1+{ t }^{ 2 } \right) \\ SQ=\sqrt { { \left( a-\dfrac { a }{ { t }^{ 2 } } \right) }^{ 2 }+\dfrac { 4{ a }^{ 2 } }{ { t }^{ 2 } } } =\quad a\left( \dfrac { { t }^{ 2 }+1 }{ { t }^{ 2 } } \right)$

Now we have,

$\dfrac { 1 }{ SP } +\dfrac { 1 }{ SQ } =\dfrac { 1 }{ a\left( 1+{ t }^{ 2 } \right) } +\dfrac { { t }^{ 2 } }{ a\left( 1+{ t }^{ 2 } \right) } =\dfrac { 1 }{ a } \\ \Rightarrow \dfrac { 1 }{ b } +\dfrac { 1 }{ c } =\dfrac { 1 }{ a } \\ \Rightarrow \dfrac { 1 }{ 3 } +\dfrac { 1 }{ 2 } =\dfrac { 1 }{ a } \\ \Rightarrow a=\dfrac { 6 }{ 5 }$

Equation of the circle with radius 3 and centre as the point of intersection of the lines

$2x + 3y = 5, 2x - y = 1$ is

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$x^2 + y^2 = 9$
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$x^2 + y^2 - 2x - 2y - 7 = 0$
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$x^2 + y^2 - 2x - 2y + 7 = 0$
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$x^2 + y^2 + 9 = 0$

Explanation

radius

$= 3$ ,

$2x+3y=5,2x-y=1$

$2x+3y=5$

$2x-y=1$

___________

$4y=4 \Rightarrow y=1$

$2x=2\Rightarrow x=1$

Center of circle

$= (1,1)$

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

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

$\Rightarrow x^{2}+y^{2}-2x-2y-7=0$

A square is inscribed in the circle

$x^2 +y^2 - 4x - 6y-5 =0$ whose sides are parallel to co-ordinate axes then vertices of square are

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$(5, 0), (5, 6), (-1, 0), (-1, 6)$
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$(5, 1), (5,-6), (-1, 1), (-1, 6)$
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$(5, 1), (5, 6), (-1, 0), (1, 6)$
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$(0, 5), (-6, 5), (0,-1),(6, 1)$

Explanation

$x^2+y^2-4x-6y-5=0$

$(x-2)^2-4+(y-3)^2-9-5=0$

$(x-2)^2+(y-3)^2=18$

centre

$(2,+3)$ radius

$=\sqrt{18}=3\sqrt{2}$

If a square is inscribe in circle,

$\Rightarrow$ diagonal of square =diameter

$\sqrt{2}a=2(radius)$

$2=2\times 3\sqrt{2}$

$\Rightarrow a=6$

$\therefore$ Side of the square is

$6$

Since the side are parallel to

$x$ and

$y$

the vertices one equidistant from the centre.

$(2+3,+3+3),(2-3,+3+3), (2-3,+3-3)$ and

$(2+3,+3-3)$

$\Rightarrow (5,6), (-1,6), (-1,0),(5,0)$

If the lines

$3x-4y-7=0$ and

$2x-3y-5=0$ are two diameters of a circle of area 154 square units , the equation of the circle is :

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$x^2+y^2+2x-2y-62=0$
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$x^2+y^2-2x+2y-62=0$
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$x^2+y^2-2x+2y-47=0$
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$x^2+y^2+2x-2y-47=0$

Explanation

Let us find intersection of

$3x-4y-7=0$

$2x-3y-5=0$

$\dfrac {8y+14}{3}-3y-5=0$

$-9y+8y-15+19=0$

$y=-1$

$x=1$

Area

$=l\ \pi \ r^2=154$

$\dfrac {22}{7}r^2 =154$

$r^2 =49$

$r=7$ units

Circle:

$(x-1)^2 +(y+1)^2 =49$

$x^2 +y^2-2x+2y-47=0$

$C$ is correct.

The lines

$2x - 3y = 5$ and

$3x - 4y = 7$ are two diameters of a circel of area

$154sq.$ units. Then the equation of circle is

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$(x+1)^2+(y+1)^2 = 49$
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$(x-1)^2 + (y-1)^2=-49$
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$(x-1)^2+(y+1)^2 = 49$
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$(x+1)^2+(y-1)^2 = 49$

Explanation

Circle area

$=\pi r^{2}=154\Rightarrow r=7\ sq.units$

intersection of diameter

$(2x-3y=5)\times 3\\ (3x-4y=7)\times 2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad -y=1\Rightarrow y=-1$

$2x=5-3=2\Rightarrow x=1$

eqn of circle

$\Rightarrow (x-1)^{2}+(y+1)^{2}=49$

A variable point

$P$ on the ellipse of eccentricity is joined to the foci

$S$ and

$s'$ . The eccentricity of the locus of the in cetre of the triangle

$PSS^{1}$ is

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$\sqrt {\dfrac {2e}{1+e}}$
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$\sqrt {\dfrac {e}{1+e}}$
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$\sqrt {\dfrac {1-e}{1+e}}$
0%
$\dfrac {e}{2(1+e)}$

Explanation

In centre of

$\triangle PS_1S_2$ is

$I=\left(\dfrac{2a^2e\cos\theta-a^2e(1-e\cos\theta)+a^2e(1+e\cos\theta)}{2ae+2a},\dfrac{2ae\sin\theta}{2a(1+e)}\right)$

$=\left(\dfrac{2a^2e\cos\theta+2a^2e^2\cos\theta}{2a(1+e)},\dfrac{be\sin\theta}{1+e}\right)$

$=\left(\dfrac{2a^2e\cos\theta+(1+e)}{2a(1+e)},\dfrac{be\sin\theta}{1+e}\right)=(x,y)$

$\dfrac{x}{ae}=\cos\theta,\ \sin\theta=\dfrac{(1+e)y}{be}$

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

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

1026976_1141657_ans_0d593268bfa04952a7541cedfafe9e3d.png

The circles

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

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

0%
Do not intersect
0%
Are not orthogonal
0%
Intersect orthogonally
0%
Concentric

Explanation

$S_1 = x^{2}+y^{2}+4y-4x+4 = 0$

$S_2 = x^2+y^2-4x-4y = 0$

$S_1 -S_2 = (x^2+y^2-4x+4y+4)-(x^2+y^2-4x-4y)$

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

$= 8y-4$

$\Rightarrow 8y-4 = 0 \Rightarrow y = 1/2$

from

$S_1 = x^2+y^2+4y-4x+4 = 0$

$C_1$ centre (2,-2)

$R_1$ Radius =

$\sqrt{4+4-4} = 2$

similarly

$S_2 = X^2+y^2-4x-4y = 0$

$C_2$ centre (2,2)

$R_2$ Radius

$= \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$

Distance between centre of circles,

$\left | C_1 C_2 \right | = \sqrt{(2-2)^2+(2+2)^2 } = \sqrt{0+4^2} = \sqrt{4^2} = 4$

Now

$R_1+R_2 = 2+2\sqrt{2} = 2 (1+\sqrt{2}) \neq \left | C_1 C_2\right |$

$R_1 +R_2>$

The condition for orthogonality

$= 2gg'+2ff' = c+c'$

$\Rightarrow 2(2)(2)+2(-2)(2) = 4+0$

$8-8 = 0\neq 4$

So, no circles do not intersect orthogonality,

Option (b) is correct

1169582_1139999_ans_0484c9d22e3341769d5a59d02a0f6123.jpeg

The set of values of p for which the power of a point

$(2,5)$ is negative with respect to a circle

${ x }^{ 2 }+{ y }^{ 2 }-8x-12y+p=0$ which neither touches the axes nor cuts them are

0%
(36,57)
0%
(36,47)
0%
(37,47)
0%
(16,47)

Explanation

Power is negative

$\Rightarrow S_{1} < 0$

$\Rightarrow 2^{2}+5^{2}-16- 60+p < 0$

$\Rightarrow 4+25-76+p < 0$

$\Rightarrow p < 47$

& square of x-coordinate of circle's

centre > square of radius (does not cut axes)

$\Rightarrow g^{2} > g^{2}+f^{2}- c$

$\Rightarrow f^{2}- c < 0 \Rightarrow f^{2} < C$

$\Rightarrow 36 < p$ similarly

$g^{2} < c$

$\Rightarrow 16 < p \Rightarrow p > 36$

$\Rightarrow p \in (36, 47)$

1156523_1136961_ans_836da089506b4ac2908f81695d0347bc.jpg

If focus of the parabola is

$(3,0)$ and length of latus rectum is

$8$ , then its vertex is

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

Explanation

$\textbf{Step -1: Identify the standard form of parabola involved based on given data}$

$\text{Given focus of the parabola (3, 0) lies on positive x-axis }$

$\text{Thus parabola is of the form of } y^2 = 4a(x-h)$

$\text{Now given that the length of latus rectum is }8$

$\Rightarrow 4a=8$

$\Rightarrow a=2$

$\text{Focus is at } (a+ h, 0) = (3, 0)$

$\Rightarrow a+ h = 3$

$\Rightarrow h = 1$

$\textbf{Step -2: Find the vertex of the parabola }$

$\text{For a parabola of the form } y^2 = 4a(x-h)$

$\text{Vertex is } (h,0) = (1, 0)$

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

If the equation

$ax^{2}+2(a^{2}+ab-16)xy+by^{2}2ax+2by-\sqrt[4]{2}=0$ represents a circle, the radius of the circle is

0%
$\sqrt{16 + \sqrt[4]{2}}$ .
0%
$\sqrt{24 + \sqrt[4]{2}}$ .
0%
$\sqrt{2}$
0%
$\sqrt[4]{2}$

Explanation

$a{x}^{2} + 2 \left( {a}^{2} + ab - 16 \right) xy + b {y}^{2} + 2ax + 2by - \sqrt[4]{2} = 0$

If the above equation represents the equation of circle then-

$a = b ..... \left( 1 \right)$

${a}^{2} + ab - 16 = 0$

$\Rightarrow {a}^{2} + {a}^{2} - 16 = 0$

$\Rightarrow 2{a}^{2} = 16$

$\Rightarrow {a}^{2} = 8$

$\therefore {b}^{2} = {a}^{2} = 8$

Radius of circle

$\left( r \right) = \sqrt{{a}^{2} + {b}^{2} - c}$

$\Rightarrow r = \sqrt{8 + 8 + \sqrt[4]{2}}$

$\Rightarrow r = \sqrt{16 + \sqrt[4]{2}}$

Hence the radius of the circle is

$\sqrt{16 + \sqrt[4]{2}}$ .

A circle has radius

$3$ units and its centre lies on the line

$y=x-1$ . if it passes through

$(7,3)$ , its equation

0%
$x^{2}+y^{2}-8x-14y=0$
0%
$x^{2}+y^{2}-8x-6y-6=0$
0%
$x^{2}+y^{2}-14x-12y-76=0$
0%
$x^{2}+y^{2}+14x+12x-70=0$

Explanation

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

Here

$\left(h,k\right)\rightarrow$ centre

Equation of line through centre

$k=h-1----(1)$

$\left(x,y\right)$ point on circle

So, given

$\left(x,y\right)= \left(7,3\right)$

So,

${\left(7-h\right)}^{2}+{\left(3-h+1\right)}^{2}={3}^{2}$

Using eqn

$(1)$

We get,

$28+{h}^{2}-11h=0$

$h=7$ oR

$4$

$k=6$ oR

$3$

Equation of circle corresponding

$\left(7,6\right)$

${x}^{2}+{y}^{2}-14x-12y+76=0$

Equation of circle corresponding

$\left(4,3\right)$

${x}^{2}+{y}^{2}-8x-6y+16=0$

Two rods of lengths 'a' and 'b' slide along coordinate aces such that their ends are concyclic. Locus of the centre of the circle is?

0%
$4(x^2+y^2)=a^2+b^2$
0%
$4x(x^2+y^2)=a^2-b^2$
0%
$4(x^2-y^2)=a^2-b^2$
0%
$xy=ab$

Explanation

$k^{2}+ \left( \dfrac{b}{\alpha} \right)^{2} =r^{2}$

$h^{2}+ \left( \dfrac{a}{\alpha} \right)^{2} =r^{2}$

$\Rightarrow 4 (h^{2}+k^{2})=a^{2}+b^{2}$

1379623_1140933_ans_9e06fd71395e4973a04cd11514307e54.png

The length of the diameter of the circle which touches the

$x-$ axis at the point

$(1,0)$ and passes through the point

$(2,3)$

0%
$6/5$
0%
$5/3$
0%
$10/3$
0%
$3/5$ .

Explanation

Given:-

A circle which touches the

$x$ -axxis at

$\left( 1, 0 \right)$ and also passes through the point

$\left( 2, 3 \right)$ .

To find:-

Diameter of the circle

Solution:-

Let the coordinates of centre of the circle be

$\left( h, k \right)$ and

$r$ be the radius of the circle.

Now,

Since the circle touches

$x$ -axis at

$\left( 1, 0 \right)$ .

$\therefore$ Radius of circle

$\left( r \right) = k$

Therefore, equation of circle is-

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

Given that the circle passes through points

$\left( 1, 0 \right)$ and

$\left( 2, 3 \right)$ , thus both coordinates will satisfy the equation of circle.

Therefore,

${\left( 1 - h \right)}^{2} + {k}^{2} = {k}^{2}$

$\Rightarrow {\left( h - 1 \right)}^{2} = 0$

$\Rightarrow h = 1 ..... \left( 1 \right)$

${\left( 2 - h \right)}^{2} + {\left( 3 - k \right)}^{2} = {k}^{2}$

$\Rightarrow {h}^{2} + 4 - 4h + 9 + {k}^{2} - 6k = {k}^{2}$

$\Rightarrow {h}^{2} - 4h - 6k + 13 = 0 ..... \left( 2 \right)$

From

${eq}^{n} \left( 1 \right)$ , we have

${1}^{2} - 4 \left( 1 \right) - 6k + 13 = 0$

$1 - 4 - 6k + 13 = 0$

$\Rightarrow 6k = 10$

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

Therefore,

Radius of circle

$= \cfrac{5}{3}$

$\therefore$ Diameter of circle

$= 2r = 2 \times \cfrac{5}{3} = \cfrac{10}{3}$

Thus the diameter of circle is

$\cfrac{10}{3}$ .

Hence the correct answer is

$\left( C \right) \cfrac{10}{3}$

$P$ and

$Q$ are any two points on the circle

$x^2 + y^2 = 4$ such that

$PQ$ is a diameter. If

$\alpha$ and

$\beta$ are the lengths of perpendicular from

$P$ and

$Q$ on

$x + y = 1$ then the maximum value of

$\alpha \beta$ is

0%
$\dfrac{1}{2}$
0%
$\dfrac{7}{2}$
0%
$1$
0%
$2$

The circle passing through the points

$(-1,0)$ and touching the y-axis at

$(0,2)$ also passes through the point:

0%
$(-\dfrac{3}{2},0)$
0%
$(-\dfrac{5}{2},2)$
0%
$(-\dfrac{3}{2},\dfrac{5}{2})$
0%
$(-4,0)$

Explanation

Let

$\left( h, k \right)$ be center of circle.

Circle touches the

$y$ -axis.

$\therefore$ Radius of circle

$= h$

Equation of circle-

${\left( x - h \right)}^{2} + {\left( y - k \right)}^{2} = {h}^{2} ..... \left( 1 \right)$

Since the circle passes through

$\left( 0, 2 \right)$

Therefore,

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

$\Rightarrow {\left( k - 2 \right)}^{2} = 0$

$\Rightarrow k = 2$

Given that the circle also passes through

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

Therefore,

${\left( -1 - h \right)}^{2} + {2}^{2} = {h}^{2}$

${h}^{2} + 1 + 2h + 4 = {h}^{2}$

$h = - \cfrac{5}{2}$

Substituting the value of

$h$ and

$k$ in

${eq}^{n} \left( 1 \right)$ , we get

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

Now, we can find the point from which the circle passes.

As the point

$\left( 4, 0 \right)$ satisfy the equation of circle.

Thus, the circle will also pass through the point

$\left( -4, 0 \right)$ .

Hence the correct answer is

$\left( D \right) \left( -4, 0 \right)$ .

The equation of the circle passing through

$(3, 6)$ and whose centre is

$(2, -1)$ is

0%
$x^{2}+y^{2}-4x+2y=45$
0%
$x^{2}+y^{2}-4x-2y+45=0$
0%
$x^{2}+y^{2}+4x-2y=45$
0%
$x^{2}+y^{2}-4x+2y+45=0$

Explanation

Equation of circle :

${ \left( x-2 \right) }^{ 2 }+{ \left( y-\left( -1 \right) \right) }^{ 2 }={ \left( \sqrt { { \left( 3-2 \right) }^{ 2 }+{ \left( 6-\left( -1 \right) \right) }^{ 2 } } \right) }^{ 2 }$

$\therefore { x }^{ 2 }-4x+4+{ y }^{ 2 }+2y+1={ \left( \sqrt { 1+49 } \right) }^{ 2 }$

$\therefore \boxed { { x }^{ 2 }+{ y }^{ 2 }-4x+2y=45 }$

Equation of circle.

$(6, -3)$ is the one extremity of diameter to the circle

$x^{2}+y^{2}-3x+8y-4=0$ then its other extremity is-

0%
$(3/2, -4)$
0%
$(-3, -5)$
0%
$(3, -5)$
0%
$(3, 5)$

The equation of circle with centre

$(1,2)$ and tangent

$x+y-5=0$ is

0%
${x}^{2}+{y}^{2}+2x-4y+6=0$
0%
${x}^{2}+{y}^{2}-2x-4y+3=0$
0%
${x}^{2}+{y}^{2}-2x+4y+8=0$
0%
${x}^{2}+{y}^{2}-2x-4y+8=0$

Eccentricity of an ellipse is

$\sqrt {\cfrac{2}{5}}$ and it passes through the point

$(-3,1)$ then its equation is

0%
$3{x^2} + 5{y^2} = 32$
0%
$2{x^2} + 3{y^2} = 33$
0%
$3{x^2} + 4{y^2} = 30$
0%
$2{x^2} + 3{y^2} = 34$

The equation of the circle having the lines

$y^{2}+2y+4x-2xy=0$ as its normals & passing through the point

$(2,1)$ is

0%
$x^{2}+y^{2}-2x-4y+3=0$
0%
$x^{2}+y^{2}-2x+4y+3=0$
0%
$x^{2}+y^{2}-2x+4y-5=0$
0%
$x^{2}+y^{2}+2x+4y+13=0$

The circle

${x^2} + {y^2} - 3x - 4y + 2 = 0$ cuts

$x$ -axis

0%
$(2,0),(-3,0)$
0%
$(3,0),(4,0)$
0%
$(1,0),(-1,0)$
0%
$(1,0),(2,0)$

Explanation

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

x-axis will be cut when

$y=0$

put

$y=0$

$x^{2}-3x+2=0$

$(x-2)(x-1)=0$

$x=1,2$

points

$(1,0),(2,0)$

D is correct.

What is the area enclosed by

$|x|+|y|=1$ ?

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

Explanation

$|x|+ |y|=1 \Rightarrow \begin{cases} x+y=1 & x \ge 0 , y \ge 0 \\ -x+y=1 & x < 0, y \ge 0 \\ -x-y=1 & x < 0 , y < 0 \\ x-y=1 & x \ge 0 , y < 0 \end{cases}$

Plotting on graph

Area enclosed

$=1 (1) =1 unit^{2}$

1372436_1168863_ans_02b7c62a0b18481c90a0ca8d7c444df4.png

Equation of the circle which passes through the centre of the circle

$x^{2} + y^{2} + 8x + 10y - 7 = 0$ and is concentric with the circle

$2x^{2} + 2y^{2} - 8x - 12y - 9 = 10$ is

0%
$x^{2} + y^{2} - 4x - 8y - 97 = 0$
0%
$x^{2} + y^{2} - 4x - 6y - 87 = 0$
0%
$x^{2} + y^{2} - 2x - 8y - 95 = 0$
0%
None of these

Explanation

First equation

$=x^2+y^2+8x+10y-7=0$

its centre

$=(-4, -5)$

Second equation of the circle

$=2x^2+2y^2+8x-12y-9=0$

its, centre

$=(2, 3)$

radius:

$=r^2\sqrt {(2+4)^2+(3+5)^2}$

$r^2=\sqrt {6^2+8^2}=\sqrt {36+64}$

$r^2 =100$

$r=10$

The equation of the concentric

circle

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

$x^2-4x+4+y^2-6y+9-100=0$

$x^2-4x+y^2-6y+13-100=0$

$x^2+y^2-4x-6y-87=0$

The centres of a set of circles, each of radius

$3$ , lie on the circle

${x}^{2}+{y}^{2}=25$ . The lotus of any point in the set is

0%
$4\le {x}^{2}+{y}^{2}\le 64$
0%
${x}^{2}+{y}^{2}\le 25$
0%
${x}^{2}+{y}^{2}\ge 25$
0%
$3\le {x}^{2}+{y}^{2}\le 9$

The curve described parametrically by

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

$y = {t^2} - t + 1$ represents

0%
hyperbola
0%
ellipse
0%
parabola
0%
rectangular hyperbola

Explanation

The equations

$x=t^2+t+1$ and

$y=t^2-t+1$

Subtracting

$y$ from

$x,$ we get

$x-y=2t$

$\Rightarrow$

$t=\dfrac{x-y}{2}$ ----- ( 1 )

Now, adding

$x$ and

$y$ we get,

$\Rightarrow$

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

Substituting the value of

$t$ from equation ( 1 ) in equation ( 2 ), we get

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

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

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

The general second degree equation is of the form

$ax^2+2hxy+by^2+2gx+2fy+c=0$ ----- ( 4 )

Comparing equations ( 3 ) and ( 4 ), we have

$a=1,h=-1,b=1,g=-1,f=-1$ and

$c=4$

This equation represents parabola if

$h^2-ab=0$

$h^2-ab=(-1)^2-(1)(1)=1-1=0$

$\therefore$ The equation ( 3 ) represents a parabola.

$2x - 3y = 5$ and

$3x - 4y = 7$ are the equation of

$2$ diameters of a circle whose area is

$88$ sq. units then the equation of the circle is :

0%
${x^2} + {y^2} + 2x - 2y - 47 = 0$
0%
${x^2} + {y^2} - 2x + 2y - 49 = 0$
0%
${x^2} + {y^2} - 2x + 2y + 47 = 0$
0%
${x^2} + {y^2} - 2x + 2y - 26 = 0$

If he equations of two diameters of a circle are

$2x + y = 6$ and

$3x + 2y = 4$ and the radius is

$10$ , find the equation of the circle.

0%
$x^{2}+y^{2}-16x+20y+64=0$
0%
$x^{2}-y^{2}+16x-20y+34=0$
0%
$x^{2}+y^{2}+6x-20y+64=0$
0%
None of these

Explanation

Equation of two diameters of circle are

$2x+y=6 ....... (i)$

$3x+2y=4 ...... (ii)$

We know intersection of these line will give centre of circle

$4x+2y=12$

$3x+2y=4$

$(-)$

$\overline {2x+0=16}$

$\Rightarrow x=8$

Put

$x=8$ in

$(i)$ we get

$16+y=6$

$y=-10$

centre of circle is

$(8, -10)$ and radius

$10$ equation of circle will be

$(x-8)^{2}+(y+10)^{2}=(10)^{2}$

$x^{2}-16x+64+y^{2}+20y+100=100$

$x^{2}+y^{2}-16x+20y+64=0$

The equation

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

0%
$r>1$
0%
$r>5$
0%
$2 < r< 5$
0%
$r<2$ or $r>5$

The equation of directrix of a parabola

$3 x + 4 y + 15 = 0$ and equation of tangent at vertex is

$3 x + 4 y - 5 = 0$ . Then the length of latus rectum is equal to-

0%
15
0%
14
0%
13
0%
16

Which of the following equations represents parametrically, parabolic profile ?

0%
$x=3\:cos\:t\:;\:y=4\:sin\:t$
0%
$x^2-2=-\:cos\:t\:;\:y=4\:cos^2\:\dfrac{t}{2}$
0%
$\sqrt{x}=tan\:t\:;\:\sqrt{y}=sec\:t$
0%
$x=\sqrt{1-sin\:t};\:y=sin\:\dfrac{t}{2}+cos\:\dfrac{t}{2}$

Explanation

Option

$a)$

$x=3\cos{t}\Rightarrow \cos{t}=\dfrac{x}{3}$

$y=4\sin{t}\Rightarrow\sin{t}=\dfrac{y}{4}$

We know the identity,

${\sin}^{2}{t}+{\cos}^{2}{t}=1$

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

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

$\Rightarrow \dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{16}=1$ represents an equation of ellipse.

Option

$b)$

${x}^{2}-2=-2\cos{t}$

$\Rightarrow {x}^{2}=2-2\cos{t}$

$\Rightarrow {x}^{2}=2\left(1-\cos{t}\right)$

$\Rightarrow {x}^{2}=2\left(1-\left(1-2{\sin}^{2}{\dfrac{t}{2}}\right)\right)$

$\Rightarrow {x}^{2}=4{\sin}^{2}{\dfrac{t}{2}}$

We have

$y=4{\cos}^{2}{\dfrac{t}{2}}$

$\Rightarrow {\cos}^{2}{\dfrac{t}{2}}=\dfrac{y}{4}$

We know the identity,

${\sin}^{2}{\dfrac{t}{2}}+{\cos}^{2}{\dfrac{t}{2}}=1$

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

$\Rightarrow {x}^{2}=4-y$ represents a parabolic profile.

Option

$c)$

$\sqrt{x}=\tan{t}, \sqrt{y}=\sec{t}$

We know that

${\sec}^{2}{t}-{\tan}^{2}{t}=1$

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

$\Rightarrow y-x=1$ or

$x-y+1=0$ is the equation of a straight line.

Option

$d)$

$x=\sqrt{1-\sin{t}}=\sqrt{{\sin}^{2}{t}+{\cos}^{2}{t}-\sin{t}}$ using the identity

${\sin}^{2}{t}+{\cos}^{2}{t}=1$

$=\sqrt{{\sin}^{2}{t}+{\cos}^{2}{t}-2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}$ using multiple angle formula

$\sin{t}=2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}$

$=\sqrt{{\left(\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}\right)}^{2}}$

$=\left|\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}\right|$

$\therefore x=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}$ and

$x=\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}$

We have

$y=\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}$

$\Rightarrow x+y=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}=2\sin{\dfrac{t}{2}}$

$\Rightarrow \sin{\dfrac{t}{2}}=\dfrac{y+x}{2}$

$\Rightarrow x-y=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}=-2\cos{\dfrac{t}{2}}$

$\Rightarrow \cos{\dfrac{t}{2}}=\dfrac{y-x}{2}$

We have the identity

${\sin}^{2}{\dfrac{t}{2}}+{\cos}^{2}{\dfrac{t}{2}}=1$

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

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

$\Rightarrow {x}^{2}+{y}^{2}=2$ is the equation of the circle.

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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