MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 2

The equation of the image of the circle

$x^{2}+y^{2}-6x-4y+12=0$ by the line mirror

$x+y-1=0$ is

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

Explanation

The image of circle w.r.t same line means image of centre w.r.t that line without changing radius
$x^{2}+y^{2}-6x-4y+12=0$
Centre $=(3,2)$ Radius $=1$
Image of $(3,2)$ w.r.t $x+y-1=0$
$\dfrac {x-3}{1}=\dfrac {y-2}{1}=-2\dfrac {(3+2-1)}{1^{1}+1^{2}}=-4$
$x=-1$ , $y=-2$
Then equation of image of circle is
$(x+1)^{2}+(y+2)^{2}=(1)^{2}$
$\Rightarrow x^{2}+y^{2}+2x+4y+4=0$

The locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle ABC is constant is a circle
whose centre is at the

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centroid of triangle ABC
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Circumcentre of triangle ABC
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Orthocentre of triangle ABC
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incentre of triangle ABC

Explanation

$AP^{2}+AB^{2}+PC^{2}=C$

$(x-x_{1})^{2}+(y-y_{1})^{2}+(x-x_{2})^{2}-(y-y_{2})^{2}+(x-x_{3})^{2}+(y-y_{3})^{2}=c$

$3x^{2}-(2x_{1}+2x_{2}+2x_{3})x+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+3y^{2}-2(y_{1}+y_{2}+y_{3})+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=c$

$x^{2}+y^{2}-\frac{2}{3}(x_{1}+x_{2}+x_{3})x-\frac{2}{3}(y_{1}+y_{2}+y_{3})c+(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-c)=0$
So, center of this circle is

$(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}),$
centroid of triangle is a center of that circle.

The lines 2x-3y

$=5$ and 3x-4y

$=7$ diameters of a circle having area as

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

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$x^{2}+y^{2}-2x+2y=62$
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$x^{2}+y^{2}+2x+2y=62$
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$x^{2}+y^{2}+2x-2y=47$
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$x^{2}+y^{2}-2x+2y=47$

Explanation

$2x-3y=5----(1)$

$3x-4y=7----(2)$
are diameter of circle so, point of intersection is centre of circle.
from (1) & (2),

$x=1, y=-1$

$\therefore$ area of circle is

$154$

$=\pi r^{2}=154$

$r^{2}=\dfrac{7}{22}\times154$

$=\dfrac{7\times14}{2}$

$r^{2}=49$

$r=7$

$\therefore$ Equation of circle with centre

$(1,-1)$ and radius

$7$ units

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

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

lf the line

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

${x}$ -axis,

$y$ -axis respectively at

${A}$ and

${B}$ , then the equation of the circle with radius

${A}{B}$ and Centre at

${A}$ is

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

Explanation

Given that:

Equation of the straight line:

$3x-2y+6=0$

Since,

$A$ is the centre and

$B$ is one end-point.

So,

$A$ is the mid-point of two diameter end points.

One point is

$B$ and let other be

$(x,y)$

So,

$-2=\dfrac{x+0}{2}, 0=\dfrac{y+3}{2}$

So,Co-ordinates of other end-point

$=\left(-4,-3\right)$

Equation of the circle as centre

$A$ is

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

or,

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

or,

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

Hence, B is the correct option.

lf the lines

$2x+3y+1=0$ and

$3x-y-4=0$ lie along diameters of a circle of circumference

$10\pi$ , then the equation of the circle is:

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$x^{2}+y^{2}-2x+2y-23=0$
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$x^{2}+y^{2}-2x-2y-23=0$
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$x^{2}+y^{2}+2x+2y-23=0$
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$x^{2}+y^{2}+2x-2y-23=0$

Explanation

$2x+3y+1=0$ and

$3x-y-4=0$ lie along diameter of circle.

So, point of intersection of

$2x+3y+1=0$ and

$3x-7-4=0$ is a centre of required circle.

$\therefore 11x-11=0.$

$x=1$ &

$y=-1.$

So, equation of circle having radius

$r=\dfrac{10}{2\pi}=5$ and centre

$(1,-1)$ is given by

$(x-1)^{2}+(y+1)^{2}=25=r^{2}$

$\Rightarrow x^{2}+y^{2}-2x+2y-23=0$

lf the lines

$2x-3y=5$ and

$3x-4y=7$ are two diameters of a circle of radius

$7$ then the equation of the circle is

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$x^{2}+y^{2}+2x-4y-47=0$
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$x^{2}+y^{2}=49$
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$x^{2}+y^{2}-2x+2y-47=0$
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$x^{2}+y^{2}=17$

Explanation

$2x-3y=5----(1)$

$3x-4y=7----(2)$
Intersection of this lines given centre of circle

$\therefore$ from (1) & (2),

$n=1, y=-1$
So, equation of circle is

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

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

A line is at a constant distance

$c$ from the origin and meets the coordinates axes in

$A$ and

$B$ . The locus of the centre of the circle passing through

$O, A, B$ is

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

Explanation

Let the equation of line be

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$
where

$a$ and

$b$ are the x-intercept and y-intercept.
Then the coordinates of A and B are

$(a,0)$ and

$(0,b)$
Distance of origin to the line is

$c$

$\Rightarrow c=\displaystyle \frac{|-1|}{\sqrt{(\dfrac{1}{a})^2+(\dfrac{1}{b})^2}}$

$\Rightarrow \displaystyle \frac{1}{c}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$

$\Rightarrow \displaystyle \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$ ....(1)
Let the center of circle through O, A, B be

$(h,k)$
Points

$O(0,0), A(a,0),(0,b)$ forms a right triangle.
So, the center of circle is the mid-point of AB i.e.

$\displaystyle(\frac{a}{2},\frac{b}{2})$

$\Rightarrow h=\displaystyle \frac{a}{2}, k=\frac{b}{2}$

$\Rightarrow a=2h, b=2k$
Substitute this value in (1), we get

$\displaystyle \frac{1}{c^2}=\frac{1}{4h^2}+\frac{1}{4k^2}$

$\Rightarrow \displaystyle \frac{4}{c^2}=\frac{1}{h^2}+\frac{1}{k^2}$

$\Rightarrow 4c^{-2}=x^{-2}+y^{-2}$ (Replacing

$h,k$ by

$x,y$ )

The area of a circle centered at

$(1,2)$ and passing through

$(4,6)$ is

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$5\pi$ sq. units
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$15\pi$ sq. units
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$25\pi$ sq. units
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$30\pi$ sq. units

Explanation

Coordinates of the centre of the circle

$\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 1,2 \right)$

and coordinates of the point through which it passes

$\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 4,6 \right)$ .

We know that the radius of the circle

$r=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 6-2 \right) }^{ 2 } } =\sqrt { 9+16 } =5$

Therefore, the area of the circle

$\pi { r }^{ 2 }=\pi { \left( 5 \right) }^{ 2 }=25\pi$ sq. units

The equation of the circle passing through the point

$(-1,2)$ and having two diameters along the pair of lines

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

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

Explanation

$x^{2}-y^{2}-4x+4y+3=0$ are pair of lines along the diameters of required circle.
So, point of intersection this pair of straight line is centre of required circle.
For point of intersection
Partial differential w.r.t

$x$ ;

$2x-4=0$ --------(1)
Partial diff. equation w rt

$y$ ;

$-2y+2=0$ -------(2)
from (1) & (2),

$x=2 y=1$

$\therefore$ Equation of circle,

$(x-2)^{2}+(y-1)^{2}=r^{2}$ .
So, (-1,2) lies on

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

$9+1=r^{2}$

$r^{2}=10$

$\therefore x^{2}-4x+y^{2}-2y+5=10$

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

The eccentricity of the conic represented by

$\sqrt{(x+2)^{2}+y^{2}}+\sqrt{(x-2)^{2}+y^{2}}=8$ is

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$\displaystyle \frac{1}{3}$
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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{4}$
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$\displaystyle \frac{1}{5}$

Explanation

This is of the form

$SP + SP^{'} = 2a$

$\therefore PP^{'} = 2ae$

$=\sqrt{(2+2)^{2}}$

$= 4$

$2ae = 4$
But

$2a = 8$ (from the equation given)

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

The total number of real tangents that can be drawn to the ellipse

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

$25x^{2}+9y^{2}=450$ passing through

$(3,5)$ is

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

Explanation

$(3,5)$ lies on

$25x^2+9y^2=450$

Therefore, one tangent can be drawn

and

$(3,5)$ lies outside

$3x^2+5y^2=32$ because

$S_1>0$

Therefore, two tangents can be drawn.

So total 3 tangents

The centre, vertex, focus of a conic are

$(0,0), (0,5), (0,6)$ . Its length of latus rectum is

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$\dfrac{11}5$
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$\dfrac75$
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$\dfrac{14}5$
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$\dfrac{22}5$

Explanation

Given centre vertex and focus of the conic are

$C(0,0), A(0,5)$ and

$(0,6)$ respectively

$\Rightarrow a=5, ae=6 \Rightarrow e=\cfrac{6}{5}$
Thus

$b^2=a^2(e^2-1)=11$
Therefore, length of latus rectum is

$=\cfrac{2b^2}{a}=\cfrac{22}{5}$
Hence, option 'D' is correct

The length of the latus rectum of the hyperbola

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

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$2$
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$1$
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$4$
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$3$

Explanation

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

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

$\Rightarrow a=2, b=1$
Thus length of the latus rectum is

$=\cfrac{2b^2}{a}=1$
Hence, option 'B' is correct

A straight line is drawn through the centre of the circle

${x}^{2}+{y}^{2}=2ax$ parallel to

${x}+2y=0$ and intersecting the circle at

${A}$ and

${B}$ . Then, the area of

$\Delta A{O}{B}$ is

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$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{5}}$
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$\displaystyle \dfrac{\mathrm{a}^{3}}{\sqrt{5}}$
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$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{34}}$
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$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{3}}$

Explanation

Given equation is $x^{2}+y^{2}=2ax$

Centre $=(a,0)$ and radius $=a$

Therefore, equation of line having slope $=$ $\dfrac {-1}{2}$ and passing through $(4,0)$
$y=\dfrac {-1}{2}(x-a)$
Line $AB$ : $2y+x=a$ -(1)
$y=\dfrac {a+x}{2}$
$d=\left | \dfrac {a}{\sqrt{5}} \right |$
So, area of $\triangle AOB =\dfrac {1}{2}\times\ AB\ \times$ (perpendicular distance of $Q$ from $AB$ )
$=\dfrac {1}{2}\times 2a\times \dfrac {a}{\sqrt{5}}$
$=\dfrac {a^{2}}{\sqrt{5}}$

The length of latus rectum of the hyperbola

$4x^{2}-9y^{2}-16x-54y-101=0$ is

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$\dfrac85$
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$\dfrac87$
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$\dfrac89$
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$\dfrac83$

Explanation

$4x^{2}-9y^{2}-16x-54y-101=0$

$\Rightarrow 4(x^2-4x)-9(y^2+6y)=101$

$\Rightarrow 4(x^2-4x+4)-9(y^2+6y+9)=101+16-81=36$

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

$\Rightarrow a=3, b=2$
Thus length of latus rectum of the given hyperbola is

$=\cfrac{2b^2}{a}=\cfrac{8}{3}$

The equation

$\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x+2)^{2}+y^{2}}=5$ represents

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a circle
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ellipse
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line segment
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an empty set

Explanation

let

$A(2,0)$ ,

$B(-2,0)$ and

$P(x,y)$ be three points

$AB=4$
Given that,

$\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x+2)^{2}+y^{2}}=5>AB$

$\Rightarrow PA+PB =$ constant

$>AB$
Therefore, locus of

$P$ is an ellipse.

The equation of directrix and latus rectum of a parabola are

$3x-4y+27=0$ and

$3x-4y+2=0$ . Then the length of latus rectum is

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$5$
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$10$
0%
$15$
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$20$

Explanation

$d=\left | \dfrac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}} \right |$

where $d$ is the distance between lines whose equations are $ax+by+C_{1}=0$ & $ax+by+C_{2}=0$

$d=\left | \dfrac{27-2}{\sqrt{4^{2}+3^2}} \right |$
$=5$
$d=5$
If the distance between vertex and latus rectum $=$ distance of vertex from directrix $=a$

then $d=2a=5$
$\Rightarrow$ Length of latus rectum $=4a=10$

A circle of radius

$5$ units passes through the points

$(7,1),(9,5)$ . If the ordinate of the centre is less than

$2$ , then the equation of the circle is

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$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}-10\mathrm{y}+16=0$
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$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}+10\mathrm{y}+16=0$
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$\mathrm{x}^{2}+\mathrm{y}^{2}-24\mathrm{x}-2\mathrm{y}+120=0$
0%
$\mathrm{x}^{2}+\mathrm{y}^{2}+24\mathrm{x}-2\mathrm{y}-120=0$

Explanation

Let

$(h, k)$ is a centre of circle , so

$(h-7)^2+(k-1)^2= 25$
and

$(h-9)^2+ (k-5)^2= 25$

$Let x^2+y^2+2gx+2fy+c= 0$ is circle .
so,

$(7, 1)$ lies on circle so,

$50+14g+2f+c= 0$ ----(1)

$(9, 5)$ lies on circle

$106+18g+10f+c= 0$ ----(2)

and

$g^2+f^2-c= 25$ -----(3)
From (1) & (2),

$4g+8f+56= 0$

$g+2f+14= 0$
From (1), (2), (3), (4),

$g= -12$ and

$f= -1, c= 120$

so,

$eq^n$ of circle is

$x^2+y^2-24x-2y+120= 0$

The equation

$(\mathrm{x}^{2}-\mathrm{a}^{2})^{2}+(\mathrm{y}^{2}-\mathrm{b}^{2})^{2}=0$ represent points which are

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collinear
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on a circle centre $(a,b)$
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on a circle centre $(0,0)$
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coincident

Explanation

$(x^{2}-a^{2})^{2}+(y^{2}-b^{2})^{2}= 0$ represents points

$x= ^{+}_{-}a$ and

$y= ^{+}_{-}b$

$\therefore (a,b), (a,-b), (-a,b)$ and

$(-a,-b)$ are four points
These point lie on a circle whose centre is at

$(0,0)$ and radius is

$\sqrt{a^2+b^2}$

Assertion(A): The difference of the focal distances of any point on the hyperbola

$\displaystyle \frac{x^{2}}{36}-\frac{y^{2}}{9}=1$ is 12.
Reason(R): The difference of the focal distances of any point on the hyperbola is equal to the length of it transverse axis

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Both A and R are true and R is the correct

explanation of A.
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Both A and R are true but R is not the correct

explanation of A.
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A is true but R is false.
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A is false but R is true.

Explanation

Clearly,

$|SP-S'P|=2a=12$
Thus statement 1 is correct.
Also statement 2 is correct and followed by statement 1.

A parabola with axis parallel to

$x$ axis passes through

$(0, 0), (2, 1), (4, -1).$ Its length of latus rectum is

0%
$\displaystyle \frac{2}{3}$
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$\displaystyle \frac{1}{4}$
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$\displaystyle \frac{7}{3}$
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$\displaystyle \frac{1}{3}$

Explanation

$(y-k)^{2}= 4a(x-h)$

It passes through

$(0,0)$

$k^{2}= -4ah$ ---(1)

It passes through

$(2,1)$

$(k-1)^{2}= 4a(2-h)$ ---(2)

It passes through

$(4,-1)$

$(k+1)^{2}= 4a(4-h)$ ---(3)
(2) + (3)

$2(k^{2}+1)= 4a(6-2h)$

$2k^{2}+2= 24a+2k^{2}$ ( form eq (1) )

$4a = \dfrac{1}{3}$

$\therefore$ length of LR

$= \dfrac{1}{3}$

Length of the latusrectum of the hyperbola

$xy=c$ ,is equal to

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$2c$
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$\sqrt{2}c$
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$2\sqrt{2}c$
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$4c$

Explanation

if co-ordinate axis is rotated by

$45^0$

Then new equation of hyperbola is

$(\dfrac{X-Y}{\sqrt2})(\dfrac{X+Y}{\sqrt2})=c$

$\Rightarrow X^2-Y^2=2c$

Therefore, length of latus rectum is

$2\dfrac{b^2}{a}=\dfrac{4c}{\sqrt{2c}}=2\sqrt2 c$

The difference between the length

$2a$ of the transverse axis of a hyperbola of eccentricity

$e$ and the length of its latus rectum is :

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$2a\left| 3-{ e }^{ 2 } \right|$
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$2a\left| 2-{ e }^{ 2 } \right|$
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$2a\left( { e }^{ 2 }-1 \right)$
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$a\left( 2{ e }^{ 2 }-1 \right)$

Explanation

Let the equation of hyperbola be

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

Length of transverse axis is

$2a$ and

Length of latus rectum is

$\displaystyle \frac { 2{ b }^{ 2 } }{ a }$

Now, difference

$\displaystyle =\left| 2a-\frac { 2{ b }^{ 2 } }{ a } \right| =\frac { 2 }{ a } \left| 2{ a }^{ 2 }-{ a }^{ 2 }{ e }^{ 2 } \right|$

$\therefore$ Difference

$=2a\left| 2-{ e }^{ 2 } \right|$

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

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

$y=\dfrac{u^{2}}{2g}$ , is

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

Explanation

$F= \left ( \dfrac{u^{2}}{2g}\sin \ 2\alpha ,\dfrac{-u^{2}}{2g}\cos \ 2\alpha \right )$

For the standard form of a parabola ,

Distance between the directrix and the focus

$\times 2 =$ Length of the latus rectum.

$L.R.=2\left(\dfrac{u^{2}}{2g}+\dfrac{u^{2}}{2g}\cos \ 2\alpha\right)$

$=\dfrac{u^{2}}{g}(1+\cos \ 2\alpha)$

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

Hence, option

$D$ is correct.

A parabola has x- axis as its axis, y- axis as its directrix and

$4a$ as its latus rectum. If the focus lies to the left side of the directrix then the equation of the parabola is

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

Explanation

Let

$P(x,y)$ be a point on the parabola
Given that the directrix is y-axis
So, coordinates of any point

$M$ on directrix is of the form

$(0,y)$
Length of latus rectum

$=4a=2(2a)$
Since, axis of parabola is x-axis, and focus is on left side of directrix
So, focus is at

$(-2a,0)$

By definition of parabola,

$PS=PM$

$(x+2a)^2+y^2=x^2$

$\Rightarrow y^2=-4ax-4a^2$

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

$L_{1}L_{2}$ is the latusrectum of

$y^{2}=12x$ ,

$P$ is any point on the directrix then the area of

$\Delta PL_{1}L_{2}$ =

0%
$32$
0%
$18$
0%
$36$
0%
$16$

Explanation

The length of latus rectum is

$2l=4a=12$

and distance between latus rectum and directrix is

$2a=6$

Therefore,

$\triangle PL_1L_2=\dfrac{(2a)(2l)}{2}=36$

The length of latus rectum of the parabola

$(x-2a)^{2}+y^{2}=x^{2}$ is

0%
$2a$
0%
$3a$
0%
$6a$
0%
$4a$

Explanation

We have,

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

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

$\Rightarrow y^2= 4ax-4a^2=4a(x-a)$
Comparing it with standard parabola

$Y^2=4bX$

$Y=y, X=x-a, b = a$
We know length of latus rectum of parabola

$Y^2=4bX$ is

$4b$
Hence length of latus rectum of given parabola is

$=4\times a = 4a$

The length of latus rectum of the parabola

$y^{2}+8x-2y+17=0$ is

0%
$2$
0%
$4$
0%
$8$
0%
$16$

Explanation

The given parabola is,

$y^{2}+8x-2y+17=0$

$\Rightarrow (y^2-2y+1)=-8x-17+1=-8x-16$

$\Rightarrow (y-1)^2=-8(x+2)$
Comparing with standard parabola

$Y^2=-4aX$

$Y=y-1,X=x+2, a=2$
Hence length of latus rectum is

$=\displaystyle 4a = 4\times 2=8$

The length of latus rectum of the hyperbola

$xy-3x-3y+7=0$ is

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

Explanation

Given Equation of Hyperbola is

$xy-3y-3x+7=0$

We can rewrite the equation as

$x(y-3)-3y+7=0$ .....

$( 1 )$

Now by adding and subtracting

$9$ in equation

$( 1 )$ , we get

$\Rightarrow$

$x(y-3)-3y+9-9+7=0$

Now eq. of Hyperbola can be written in the simple terms as

$x(y-3)-3(y-3)-2=0$

$\Rightarrow$

$(x-3)(y-3)=2$ or

$(x-3)(y-3)=(\sqrt{2})^2$ ......

$( 2 )$

Equation

$( 2 )$ is similar to equation of a rectangular Hyperbola of the form

$xy=c^2$ , with shifted origin at

$(3,3)$

So given Hyperbola is also a rectangular Hyperbola, with

$c=\sqrt{2}$

We know that for a rectangular Hyperbola

$b=a=c\sqrt{2}$

So value of

$a$ for given Hyperbola

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

$\Rightarrow$ For any rectangular Hyperbola length of latusrectum

$=2a$

Hence, the length of latusrectum of given Hyperbola

$=2 \times 2=4$

The equation of the circle having centre

$(1,\ -2)$ and passing through the point of intersection of the lines

$3x+y=14$ and

$2x+5y=18$ is

0%
$x^2+y^2-2x+4y-20=0$
0%
$x^2+y^2-2x-4y-20=0$
0%
$x^2+y^2+2x-4y-20=0$
0%
$x^2+y^2+2x+4y-20=0$

Explanation

The circle passes through intersection of

$3x+y=14$ and

$2x+5y=18$ .

$\therefore 5(3x+y-14)-2x-5y+18=0\implies 15x-70-2x+18=0\implies x=4\implies y=2$

Hence, equation of circle is

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

$\therefore x^2-2x+1+y^2+4y+4=25$

$\therefore x^2+y^2-2x+4y-20=0$

This is the required equation.

The centre of a circle is

$(2a, a-7$ ). Find the values of

$a$ if the circle passes through the point (11, -9) and has diameter

$10\sqrt{2}$ units.

0%
$\dfrac 35$
0%
$3$
0%
$5$
0%
$\dfrac 53$

Explanation

$Let\quad the\quad given\quad point\quad P\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 11,-9 \right) \quad lie\quad on\quad a\quad circle\quad with\quad centre\quad at\quad$

$C\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 2a,a-7 \right) \quad \& \quad radius=r.$

$Then\quad PC=r.......(i)\quad$

$Given\quad that\quad the\quad diameter\quad of\quad the\quad circle=10\sqrt { 2 } units.$

$\therefore \quad r=\dfrac { diameter }{ 2 } =\dfrac { 10\sqrt { 2 } units }{ 2 } =5\sqrt { 2 } units....(ii).$

$Again,\quad by\quad distance\quad formula,\quad d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } \quad$

$\therefore \quad PC=d=\sqrt { { \left( 11-2a \right) }^{ 2 }+{ \left( -9-a+7 \right) }^{ 2 } } =\sqrt { 5{ a }^{ 2 }-40a+125 } .$

$\therefore \quad From\quad (i)\quad \& \quad (ii)\quad we\quad get$

$5{ a }^{ 2 }-40a+125=({ 5\sqrt { 2 } ) }^{ 2 }$

$\Longrightarrow 5{ a }^{ 2 }-40a+75=0$

$\Longrightarrow \left( a-5 \right) \left( 5a-3 \right) =0$

$\Longrightarrow a=5,\dfrac { 3 }{ 5 }$

$Ans\quad a=5,\dfrac { 3 }{ 5 }$

The equation of parabola whose latus rectum is

$2$ units, axis is

$x+y-2=0$ and tangent at the vertex is

$x-y+4=0$ is given by

0%
$(x+y-2)^{2} = 4\sqrt{2}(x-y+4)^{2}$
0%
$(x-y-4)^{2} = 4\sqrt{2}(x+y-2)$
0%
$(x+y-2)^{2} = 2\sqrt{2}(x-y+4)$
0%
$(x-y-4)^{2} =2\sqrt{2}(x-y+2)^{2}$

Explanation

${\textbf{Step - 1: Find value of 'a' from Length of latus rectum,}}$

${\text{We know that length of latus rectum of any parabola is 4a.}}$

${\text{Given that 4a = 2 }} \Rightarrow {\text{a = }}\dfrac{1}{2}.$

${\text{Also given that the axis is x + y = 2}}.$

${\textbf{Step - 2: Use relation between Perpendicular distance from axis and Distance from tangent of a parabola.}}$

${\text{From parabola's formula we know that,}}$

$\Rightarrow {\left( {{\text{perpendicular distance from axis}}} \right)^2}{\text{ = 4a}}\left( {{\text{distance from tangent}}} \right).$

$\Rightarrow {\text{ }}{\left( {\dfrac{{{\text{x + y}} - {\text{2}}}}{{\sqrt 2 }}} \right)^2} = 4\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{{\text{x}} - {\text{y + 4}}}}{{\sqrt 2 }}} \right).$

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

${\textbf{Hence, The equation is }}{\left( {{\textbf{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\textbf{x}} - {\textbf{y + 4}}} \right).(C)$

$(4-\mathrm{a})\mathrm{x}^{2}+(12-\mathrm{a})\mathrm{y}^{2}=\mathrm{a}^{2}-16\mathrm{a}+48$ represents an ellipse. Then:

0%
$a<12$
0%
$a<4$
0%
$a>4$ & $a<12$
0%
$a>0$

Explanation

LHS of equation,we can rewrite it as

$\displaystyle { a }^{ 2 }-16a+48=(a-4)(a-12)=(4-a)(12-a)$

Therefore the given equation can be written as

$\displaystyle \frac { { x }^{ 2 } }{ 12-a } +\frac { { y }^{ 2 } }{ 4-a } =1$

This will represent an ellipse ,If

$\displaystyle 4-a>0\Longrightarrow a<4$

(

$\displaystyle \ddot { . } 4-a$ is the smaller of

$\displaystyle 12-a$ )

So, taking intersection of

$\displaystyle 12-a$ and

$\displaystyle 12-a>0.$

Latus rectum of a parabola is a ........ line segment with respect to the axis of the parabola through the focus whose endpoints lie on the parabola.

0%
perpendicular
0%
parallel
0%
tilted
0%
None of these

Explanation

Consider the above image which shows that the latus rectum of a parabola is a line segment

$perpendicular$ to the axis of the parabola, through the focus and whose end points lie on the parabola.

Which of the following is/are not false?

0%
The mid point of the line segment joining the foci is called the centre of the ellipse.
0%
The line segment through the foci of the ellipse is called the major axis.
0%
The end points of the major axis are called the vertices of the ellipse.
0%
Ellipse is symmetric with respect to Y-axis only.

The arrangement of the following parabolas in the ascending order of their length of latusrectum
A)

$y=4x^{2}+x+1$ B)

$2y=x^{2}+x+5$
C)

$x=2y^{2}+y+3$ D)

$y^{2}+x+y+9=0$

0%
$B,D,C,A$
0%
$A,B,C,D$
0%
$A,C,D,B.$
0%
$A,D,C,B$

Explanation

We find the latus rectum for the given parabolas.

$y=4{ x }^{ 2 }+x+1$

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

Adding

$\displaystyle\frac { 1 }{ 64 }$ to both sides we get

$\displaystyle{ \Rightarrow \left( x+\frac { 1 }{ 8 } \right) }^{ 2 }=\frac { 1 }{ 4 } \left( y-\frac { 15 }{ 16 } \right)$

Length of latus rectum is

$\displaystyle\frac{1}{4}$

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

$\Rightarrow 2y-5={ x }^{ 2 }+x$

Adding

$\displaystyle\frac { 1 }{ 4 }$ to both sides we get

$\displaystyle{ \Rightarrow \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }=2\left( y-\frac { 19 }{ 8 } \right)$

Length of latus rectum is

$2$

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

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

Adding

$\displaystyle\frac { 1 }{ 16 }$ to both sides we get

$\displaystyle{ \Rightarrow \left( y+\frac { 1 }{ 4 } \right) }^{ 2 }=\frac { 1 }{ 2 } \left( x-\frac { 23 }{ 8 } \right)$

Length of latus rectum is

$\displaystyle\frac{1}{2}$

${ y }^{ 2 }+y+x+9=0$

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

Adding

$\displaystyle\frac { 1 }{ 4 }$ to both sides we get

$\displaystyle{ \Rightarrow \left( y+\frac { 1 }{ 2 } \right) }^{ 2 }=-1\left( x+\frac { 35 }{ 4 } \right)$

Length of latus rectum is

$1$

So the ascending order of latus rectum is

$A,C,D,B$

${a}\neq 0$ and the line

$2{b}{x}+3 cy+4{d}=0$ passes through the points of intersection of the parabolas

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

${x}^{2}=4ay,$ then

0%
${d}^{2}+(2{b}+3{c})^{2}=0$
0%
${d}^{2}+(3{b}+2{c})^{2}=0$
0%
${d}^{2}+(2{b}-3{c})^{2}=0$
0%
${d}^{2}+(3{b}-2{c})^{2}=0$

Explanation

The equation of parabolas are

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

${ x }^{ 2 }=4ay$

On solving these we get

$x=0$ and

$x=4a$

Also

$y=0$ and

$y=4a$

Therefore point of intersection of parabolas are

$A(0,0)$ and

$B(4a,4a)$

Also line

$2bx+3cy+4d=0$ passes through A and B

$\therefore d=0$ ...(1)

$2b.4a+3c.4a+4d=0\Rightarrow 2ab+3ac+d=0$

$\\ \Rightarrow a\left( 2b+3c \right) =0\Rightarrow 2b+3c=0$ ...(2)

On squaring (1) and (2) and then adding, we get

${ d }^{ 2 }+{ \left( 2b+3c \right) }^{ 2 }=0$

If the equation of the incircle of an equilateral triangle is

${ x }^{ 2 }+{ y }^{ 2 }+4x-6y+4=0$ , then the equation of the circumcircle of the triangle is

0%
${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-23=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-4x-6y-23=0$
0%
None of these

Explanation

Given equation of the incirle is

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

Its incenter is

$(-2,3)$ and inradius

$=\sqrt { 4+9-4 } =3$

Since in an equilateral triangle, the incenter and the circumcenter coincide,

$\therefore$ Circumcenter

$=(-2,3)$

Also, in an equilateral triangle, circumradius

$=2($ inradius

$)$

$\therefore$ Circumradius

$=2.3=6$

$\therefore$ The equation of the circumcircle is

${ \left( x+2 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ \left( 6 \right) }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0$

The length of the latus rectum of the parabola

$169\left \{ (x-1)^{2}+(y-3)^{2} \right \}=(5x-12y+17)^{2}$ is

0%
$\dfrac{14}{11}$
0%
$\dfrac{12}{13}$
0%
$\dfrac{28}{13}$
0%
none of these

Explanation

Given parabola may be written as,

$\displaystyle (x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{\sqrt{5^2+12^2}}\right)^2$

Thus focus of the parabola is

$(1,3)$ and directrix is

$5x-12y+17$

Now distance between directrix to the focus is

$=\cfrac{5(1)-12(3)+17}{\sqrt{5^2+12^2}}=\cfrac{14}{13}$

Hence length of latus rectum of the parabola is

$=2\times \cfrac{14}{13}=\cfrac{28}{13}$

The equation of a locus is

$y^{2}+2ax+2by+c=0$ . Then

0%
it is an ellipse
0%
it is a parabola
0%
latus rectum $=a$
0%
latus rectum $=2a$

Explanation

Given equation of locus is

$\displaystyle { y }^{ 2 }+2ax+2by+c=0$ , which can be written as

$\displaystyle { (y+b) }^{ 2 }=-4\left(\frac { a }{ 2 } \right)\left(x+\frac { c-{ b }^{ 2 } }{ 2a } \right)$ which is locus of parabola.
For parabola with equation

$\displaystyle {y}^{2}=4ax$ ,

$\displaystyle 4a$ is length of latus rectum.
Then for given locus, length of latus rectum is

$\displaystyle 4\left(\frac{a}{2}\right)=2a$ .

$\dfrac {x^2}{r^2-r-6}+\dfrac {y^2}{r^2-6r+5}=1$ will represents the ellipse, if r lies in the interval:

0%
$(-\infty, -2)$
0%
$(3, \infty)$
0%
$(5, \infty)$
0%
$(1, \infty)$

Explanation

Given, Equation of ellipse as

$\dfrac{x^2}{(r^2-r-6)}+\dfrac{y^2}{(r^2-6r+5)}=1$

For

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ to be an ellipse equation

$\Rightarrow a^2>0, \ b^2>0$

$\therefore (r^2-r-6)>0, \ \ \ (r^2-6r+5)>0$

$\Rightarrow (r-3)(r+2)>0, \ \ \ (r-1)(r-5)>0$

$\Rightarrow r\in (-\infty,-2) \cup (3,\infty)\;$ and

$\;r\in (-\infty,1) \cup (5,\infty)$

$\therefore r\in(-\infty,-2) \cup (5,\infty)$

Options

$A,C$ are true.

The equation

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

0%
a point
0%
a pair of straight lines
0%
a circle of non zero radius
0%
none of these

Explanation

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

$(x-1)^{2}+(y+2)^{2} -5+5=0$

$\Rightarrow (x-1)^{2}+(y+2)^{2}=0$
Since, radius is

$0$ , hence its a point

Alternative method:
Here,

$a=b=1$

$r=\sqrt{1+4-5}=0$
Hence, a circle of radius

$0$ . So, its a point.

If major axis is the x-axis and passes through the points

$(4, 3)$ and

$(6, 2)$ , then the equation for the ellipse whose centre is the origin is satisfies the given condition.

0%
$\displaystyle \frac{x^2}{52} + \frac{y^2}{13} =1$
0%
$\displaystyle \frac{x^2}{13} + \frac{y^2}{52} =1$
0%
$\displaystyle \frac{x^2}{52} - \frac{y^2}{13} =1$
0%
$\displaystyle \frac{x^2}{13} - \frac{y^2}{52} =0$

Explanation

Major axis is x - axis
Let the equation of the ellipse be

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

$(4,3)$ and

$(6, 2)$ lies on it

$\displaystyle \frac{16}{a^2} + \frac{9}{b^2} =1$ ........

$(i)$

$\displaystyle \frac{36}{a^2} + \frac{4}{b^2} =1$ ........

$(ii)$
Subtracting

$(ii)$ from

$(i)$ , we get

$\displaystyle \frac{-20}{a^2} + \frac{5}{b^2} =0$

$\Rightarrow 5a^2 = 20 b^2 \Rightarrow a^2 = 4b^2$
Putting the value of

$a^2$ in eqn.

$(i)$

$\displaystyle \frac{16}{4b^2} + \frac{9}{b^2} = 1$

$\therefore b^2 =13$ and

$a^2 = 4b^2 = 4 \times 13 =52$

$\therefore$ Equation of ellipse is

$\displaystyle \frac{x^2 }{52}+\frac{y^2}{13}=1$

The equation

$7y^2-9x^2+54x-28y-116=0$ represents

0%
a hyperbola
0%
a parabola
0%
an ellipse
0%
a pair of straight lines

Explanation

Given conic is

$7y^2-9x^2+54x-28y-116=0$

Here

$a=-9, b=7, c=-116, f=-14, g=27, h=0$

$\therefore \Delta=abc+2fgh-af^2-bg^2-ch^2=-9.7.(-116)+0-9(14)^2-7(27)^2-0=441\neq 0$

and

$h^2=0, ab=-63$

clearly

$h^2> ab$

we know that for a equation

$ax^2+by^2+2hxy2gx+2fy+c=0$ it represents a hyperbola if

$Δ ≠ 0$ and

$h² > ab$

Hence given conic represent hyperbola.

If the parabola

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

$(3,\:2)$ then the length of latus rectum is

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{2}{3}$
0%
$1$
0%
$\displaystyle \frac{4}{3}$

Explanation

If the parabola

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

$(3,\:2)$
therefore,

$4=4a(3)$

$\Rightarrow a=\dfrac{1}{3}$
Therefore, length of latus rectum:

$l=4a=\dfrac{4}{3}$

The equation of the circle passing through the point

$(1, 1)$ and having two diameters along the pair of lines

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

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

Explanation

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

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

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

Thus lines are

$x-y=-1$ and

$x+y=3$
Solving these lines we get the centre of the required circle which is

$(1, 2)$

Thus radius of the circle is

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

Hence required circle is

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

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

$\left ( 3,3 \right )$ and directrix is

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

0%
$1$
0%
$2$
0%
$4$
0%
$8$

Explanation

Length of the latus rectum

$(l)=2($ perpendicular distance from focus to directix

$)$

$l=2\left| \dfrac { 3\left( 3 \right) -4\left( 3 \right) -2 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right|=2$

The triangle PQR is inscribed in the circle

$\displaystyle x^{2}+y^{2}= 25.$ If

$Q$ and

$R$ have coordinates

$\displaystyle \left ( 3, 4 \right )$ and

$\displaystyle \left ( -4, 3 \right ),$ respectively, then

$\displaystyle \angle QPR$ is equal to

0%
$\displaystyle \frac{\pi }{2}$
0%
$\displaystyle \frac{\pi }{3}$
0%
$\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \frac{\pi }{6}$

Explanation

Now, length of chord

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

Also,

$OR=OQ=5$ (radii of the circle)

$\therefore \angle ROQ=90^{\circ}$

Using the theorem that arc subtended by a chord is twice the angle subtended by the chord at the arc of the circle,

we have

$\angle QPR=45^{\circ}=\dfrac{\pi}{4}$

The lines

$\displaystyle 2x - 3y = 5$ and

$\displaystyle 3x - 4y = 7$ intersect at the center of the circle whose area is

$154$ sq. units, then equation of circle is

0%
$\displaystyle x^2 + y^2 - 2x + 2y = 47$
0%
$\displaystyle x^2 + y^2 + 2x - 2y = 31$
0%
$\displaystyle x^2 + y^2 - 2x - 2y = 47$
0%
$\displaystyle x^2 + y^2 - 2x - 2y = 31$

Explanation

Since, the lines

$\displaystyle 2x - 3y = 5 \dots (1)$ and

$\displaystyle 3x - 4y = 7 \dots (2)$ passes through the center of the circle.

Multiplying

$(1)$ by 4 and

$(2)$ by 3. Further subtracting we get,

$-1x=-1 \Rightarrow x=1$

Substituting

$x=1$ in

$(1)$ . We get,

$y=-1$

So, point of intersection of these two lines is

$(1,-1)$

So, center is at

$(1,-1)$

Area of circle

$=\pi r^2$

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

$r^2=49$

Equation of circle is

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

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

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

$\Rightarrow x^2+y^2-2x+2y=47$

If the centroid of an equilateral triangle is

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

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

0%
$x^{2}+y^{2}-2x-2y-3=0$
0%
$x^{2}+y^{2}+2x-2y-3=0$
0%
$x^{2}+y^{2}+2x+2y-3=0$
0%
none of these

Explanation

Given centroid of an equilateral triangle is

$G(1,1)$ .
We know that in an equilateral triangle, centroid, circumcenter and incenter are at the same point.
So, the circumcenter is at

$G(1,1)$ .
Given one vertex of equilateral triangle at

$A(-1,2)$
So, circumradius

$=AG=\sqrt { 5 }$
So, equation of circumcircle is

$(x-1)^2+(y-1)^2=5$

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

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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