MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 4

If the straight line

$y=mx+c$ is parallel to the axis of the parabola

$y^2=lx$ and intersects the parabola at

$\left(\dfrac{c^2}{8}, c\right)$ then the length of the latus rectum is

0%
$2$
0%
$3$
0%
$4$
0%
$8$

Explanation

Length of Latus rectum of the parabola

$y^2=4ax$ is

$4|a|$

Here, parabola given is

$y^2 = lx$

So, length of latus rectum will be

$l$

It is also given that a straight line

$y = mx+c$ intersect the given parabola at

$\left(\dfrac{c^2}8 ,c\right)$

Therefore, this point

$\left(\dfrac{c^2}8 ,c\right)$ should satisfy the equation of parabola

$y^2 = lx$

$\because y = c, x = \dfrac{c^2}8$

$\Rightarrow c^2 = l\times\dfrac{c^2}8$

$\Rightarrow l = 8$

$\therefore$ The length of latus rectum will be

$l = 8$

The area of the circle represented by the equation

${(x+3)}^{2}+{(y+1)}^{2}=25$ is

0%
$4\pi$
0%
$5\pi$
0%
$16\pi$
0%
$25\pi$

Explanation

The general equation of circle with centre

$(h,k)$ and radius

$r$ is given by

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

It is given that the equation of the circle is

$(x+3)^2+(y+1)^2=25$

Therefore,

$h=-3$ ,

$k=-1$ and

$r=5$ .

The area of the circle with radius

$r$ is

$\pi { r }^{ 2 }$ , thus the area of the circle with radius

$r=5$ is:

$\pi { r }^{ 2 }=\pi { (5) }^{ 2 }=25\pi$

Hence, the area of the circle is

$25\pi$ .

The radius of the circle passing through the point

$(6, 2)$ and two of whose diameters are

$\displaystyle x+y=6$ and

$\displaystyle x+2y=4$ is:

0%
$4$
0%
$6$
0%
$20$
0%
$\displaystyle \sqrt { 20 }$

Explanation

Point of intersection of the given diameters is

$(8,-2)$ which is the centre of the circle

Also the circle pass through the point

$(6,2)$ , so the radius is

$=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{20}$

Write the equation of the circle with center at

$(0,0)$ and a radius of

$6$

0%
$(x-6)^2+y^2=36$
0%
$x^2+(y-6)^2=0$
0%
$x^2+y^2=36$
0%
$x^2+y^2=-36$

Explanation

An equation of the circle with center

$(h,k)$ and radius

$r$ is

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

So, if the center is

$(0,0)$ and the radius is

$6$ , an equation of the circle is:

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

On simplifying, we get:

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

Hence, the equation of the circle is

$x^{ 2 }+y^{ 2 }=36$ .

The graph of the equation

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

0%
a circle
0%
an ellipse
0%
a hyperbola
0%
a parabola

The graph of the equation

$4y^2 + x^2= 25$ is

0%
a circle
0%
an ellipse
0%
a hyperbola
0%
a parabola
0%
a straight line

Explanation

Given,

$4{y}^{2}+{x}^{2}=25$

$\Rightarrow \dfrac { { y }^{ 2 } }{ 25/4 } +\dfrac { { x }^{ 2 } }{ 25 } =1$
It is in the form of ellipse

$\left (\dfrac { { y }^{ 2 } }{ {a}^{2} } +\dfrac { { x }^{ 2 } }{ {b}^{2} } =1\right)$
So, the correct answer is option

$B$ .

Which of the following is an equation of the circle with its center at

$(0,0)$ that passes through

$(3,4)$ in the standard

$(x,y)$ coordinate plane?

0%
$x+y=1$
0%
$x-y=25$
0%
${x}^{2}+y=25$
0%
${x}^{2}+{y}^{2}=5$
0%
${x}^{2}+{y}^{2}=25$

Explanation

Let the radius of the circle be '

$r$ '

Then the equation of the circle is

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

Hence

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

Now, it passes through

$(3,4)$ .

Hence

$3^{2}+4^{2}=r^{2}$

$\Rightarrow r^{2}=25$ .
Therefore, the equation of the circle is

$x^{2}+y^{2}=25$ .

What is the approximate radius of the circle whose equation is

$(x-\sqrt{3})^2+(y+2)^2=11$ ?

0%
1.71
0%
2.33
0%
3.32
0%
3.85
0%
4.27

Explanation

The radius of given circle is $\sqrt{11} = 3.32$

A circle with center

$(3, 8)$ contains the point

$(2, -1)$ . Another point on the circle is:

0%
$(1, -10)$
0%
$(4, 17)$
0%
$(5, -9)$
0%
$(7, 15)$
0%
$(9, 6)$

Explanation

$(x-3)^2+(y-8)^2=r^2$
Given

$(2,-1)$ lies on the circle

$(2-3)^2+(-1-8)^2=r^2$

$1+81=r^2$

$r^2=82$
Circle equation is :

$(x-3)^2+(y-8)^2=82$
By trial and error, substitute the point
in the above equation

$(4-3)^2+(17-8)^2=82$
hence ,

$(4,17)$ satisfy the circle euation.

Identify the polynomial represented by the graph?

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

Which of the following is an equation of a circle in the

$xy$ -plane with center

$\left(0, 4\right)$ and a radius with endpoint

$\left(\dfrac{4}{3}, 5\right)$ ?

0%
${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 25 }{ 9 }$
0%
${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 25 }{ 9 }$
0%
${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 5 }{ 3 }$
0%
${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 3 }{ 5 }$

Explanation

Given, center of circle is

$(0,4)$ and point on circle

$\left (\dfrac 43,5\right)$

$\therefore$ radius of circle

$=\sqrt {\left (\dfrac 43-0\right)^2+(5-4)^2}$

$= \sqrt {\dfrac {16}{9}+1}=\sqrt {\dfrac {25}9}$

Since, Equation of the circle having centre

$(x_1,y_1)$ and radius '

$r$ ' is

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

$\therefore$ Equation of given circle

$= (x-0)^2+(y-4)^2=\left (\sqrt {\dfrac {25}9}\right)^2$

$= x^2+(y-4)^2=\dfrac {25}9$

Hence, option A is correct.

The least value of

$2x^{2} + y^{2} + 2xy + 2x - 3y + 8$ for real numbers

$x$ and

$y$ is

0%
$2$
0%
$8$
0%
$3$
0%
$1$
0%
$-\dfrac12$

Explanation

$E = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$

$\Rightarrow E = \dfrac {1}{2} (4x^{2} + 2y^{2} + 4xy + 4x - 6y + 16)$

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

This happen when the

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

Which i possible when

$(x,y) \equiv \left(-\dfrac52,4\right)$

So, least value is

$-\dfrac {1}{2}$

Locus of the point

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

$x- y- 1 = 0$ is a

0%
Straight line
0%
Circle
0%
Parabola
0%
None of these

Explanation

Locus of point

$\left( \sqrt { 3h } ,\sqrt { 3k+2 } \right)$ if it lies on line

$x-y-1=0$ is :

$\sqrt { 3 } h-\sqrt { 3k+2 } -1=0\\ { (\sqrt { 3 } h-1) }^{ 2 }={ (\sqrt { 3k+2 } ) }^{ 2 }\\ { 3k }^{ 2 }+1-2\sqrt { 3 } h=3k+2\\ \left( h-\cfrac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }=3\left( k+\cfrac { 2 }{ 9 } \right) \\ x^{ 2 }=3y$

Which is parabola.

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

$(2, -3)$ and the directrix

$x = 4$ is

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

Explanation

Distance of vertex

$(2,-3)$ from directrix

$x=4$ is

$\quad \quad =\left|\dfrac{2-4}{\sqrt{1^2+0^2}} \right|=2$

So length of latue rectum of above parabola is

$=4\times$ distance of vertex to directrix

$=4\times 2=8$

The graph of the equation

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

0%
an ellipse
0%
a circle
0%
a hyperbola
0%
a parabola
0%
two straight lines

Explanation

Given,

${x}^{2}+\dfrac{{y}^{2}}{4}=1$
It is in the form of

$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$
Therefore, it represents an ellipse.

Equation of circle with center (-a, -b) and radius

$\sqrt{a^2-b^2}$ is.

0%
$x^2+y^2-2ax-2by - 2b^2=0$
0%
$x^2+y^2-2ax-2by + 2a^2=0$
0%
$x^2+y^2+2ax + 2by + 2b^2=0$
0%
$x^2+y^2-2ax-2by + 2b^2=0$

Explanation

Fact: equation of circle with centre

$(h,k)$ and radius

$r$ is given by

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

Hence equation of circle with centre

$(-a,-b)$ having radius

$\sqrt{a^2-b^2}$ is given by,

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

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

$\Rightarrow x^2+y^2+2ax+2by+2b^2=0$

Which ordered number pair represents the center of the circle

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

0%
(9,4)
0%
(3,2)
0%
(3,-2)
0%
(6,4)

Explanation

The equation of circle is

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

$\Rightarrow x^2-6x+9+y^2+4y+4=12+13$

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

Comparing above equation with

$(x-h)^2+(y-k)^2=r^2$
Therefore, we get the center of circle as

$(3,-2)$ .

The asymptotes of a hyperbola

$4x^2 - 9y^2=36$ are

0%
$2x \pm 3y = 1$
0%
$2x \pm 3y = 0$
0%
$3x \pm 2y = 1$
0%
None

Explanation

The equation of hyperbola is

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

So the equation of asymptotes is

$\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{4}=0$

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

$\Rightarrow 2x \pm 3y=0$

Therefore option

$B$ is correct

The equation of hyperbola whose coordinates of the foci are

$(\pm8,0)$ and the lenght of latus rectum is

$24$ units, is

0%
$3{ x }^{ 2 }-{ y }^{ 2 }=48$
0%
$4{ x }^{ 2 }-{ y }^{ 2 }=48$
0%
${ x }^{ 2 }-3{ y }^{ 2 }=48$
0%
${ x }^{ 2 }-4{ y }^{ 2 }=48$

Explanation

The Foci of hyperbola are

$(\pm8,0)$ , hence Foci lie on

$x$ - axis.

We know that foci of hyperbola lie at

$(\pm ae,0)$ , So

$ae = 8$ ...

$(1)$

squaring both sides of equation

$(1)$ , we get,

$\Rightarrow a^2e^2 = 64$

Eccentricity of hyperbola

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

$\Rightarrow a^2(1+\dfrac{b^2}{a^2}) = 64$

$\Rightarrow a^2 + b^2 = 64$ ...

$(2)$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola

$= \dfrac{2b^2}{a} = 24$

$\Rightarrow b^2 = 12a$ ...

$(3)$

putting value of

$b^2$ in eq.

$(2)$ , we get,

$\Rightarrow a^2 +12a -64 = 0$

Hence

$a = 4, -16$

$a$ is always taken as positive value so

$a =4$

from eq.

$(3)$ ,

$b = \sqrt{48}$

Hence equation of hyperbola is

$\dfrac{x^2}{16} -\dfrac {y^2}{48} = 1$

$3x^2 -y^2 = 48$ , So correct option is

$A$ .

The lines

$2x - 3y - 5 = 0$ and

$3x -4y = 7$ are diameters of a circle of area 154 sq units, then the equation of the circle is.( Use

$\pi = \dfrac{22}{7}$ )

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

Explanation

The centre of the required circle lies at the intersection of

$2x - 3y- 5 = 0$ and

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

Thus, the coordinates of the centre are

$(1, -1)$ .

Let

$r$ be the radius of the circle.

$\pi r^2 =154$

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

Hence, the equation of required circle is

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

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

The length of latus rectum of the ellipse

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

0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
0%
$6$
0%
$12$

Explanation

The length of the latus rectum of an ellipse

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

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

The given equation of ellipse is

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

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

Thus, the length of latus rectum will be

$\dfrac{2(2)^{2}}{3}=\dfrac{8}{3}$

Consider the parametric equation

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

What does the equation represent?

0%
It represents a circle of diameter $a$
0%
It represents a circle of radius $a$
0%
It represents a parabola
0%
None of the above

Explanation

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

Let

$t=\tan\theta$

$\Rightarrow x=a\sin 2\theta , y=a\cos 2\theta \\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$

It represents a circle of radius

$a$ .

What is the radius of the circle passing through the point

$(2, 4)$ and having centre at the intersection of the lines

$x - y = 4$ and

$2x + 3y + 7 = 0$ ?

0%
$3$ units
0%
$5$ units
0%
$3 \sqrt 3$ units
0%
$5 \sqrt 2$ units

Explanation

The line

$x-y=4$ and

$2x+3y+7=0$ intersect at the point

$(1,-3)$

So, the centre of circle lies at

$(1,-3)$

Point

$(2,4)$ lies on the circle.

So, radius = distance of

$(2,4)$ from

$(1,-3)$

$=\sqrt{(2-1)^2+(4-(-3)^2}$

$=\sqrt{1^2+7^2}$

$=\sqrt{50}$

$=5\sqrt2$ units

So, the answer is option (D)

The differential equation

$(3x + 4y + 1)dx + (4x + 5y + 1) dy = 0$ represents a family of

0%
Circles
0%
Parabolas
0%
Ellipses
0%
Hyperbolas

Explanation

The given differential equation is

$(3x + 4y + 1)dx + (4x + 5y + 1)dy = 0... (i)$
Comparing eq. (i) with

$Mdx + Ndy = 0$ ,
we get

$M = 3x + 4y + 1$
and

$N = 4x + 5y + 1$
Here,

$\dfrac {\partial M}{\partial y} = \dfrac {\partial N}{\partial x} = 4$
Hence, eq. (i) is exact and solution is given by

$\int (3x + 4y + 1)dx + \int (4x + 5y + 1)dy = C$

$\Rightarrow \dfrac {3x^{2}}{2} + 4xy + x + 4xy + \dfrac {5y^{2}}{2} + y - C = 0$

$\Rightarrow 3x^{2} + 16xy + 2x + 5y^{2} + 2y - 2C = 0$

$\Rightarrow 3x^{2} + 2\times 8xy + 2x + 5y^{2} + 2y + C' = 0 ... (ii)$
where,

$C' = -2C$
On comparing eq. (ii) with standard form of conic section

$ax^{2} + 2hxy + by^{2} + 2gx + 2fy + C = 0$
We get,

$a = 3, h = 8, b = 5$
Here,

$h^{2} - ab = 64 - 15 = 49 > 0$
Hence, the solution of differential equation represents family of hyperbolas.

The line

$(x-2)\cos \theta +(y-2)\sin \theta =1$ touches a circle for all value of

$\theta$ , then the equation of circle is

0%
$x^2+y^2-4x-4y+7=0$
0%
$x^2+y^2+4x+4y+7=0$
0%
$x^2+y^2-4x-4y-7=0$
0%
$None\ of\ the\ above$

Explanation

Given line is

$(x-2)cos \,\theta +(y-2)sin\, \theta =1$

$\Rightarrow (x-2)cos \theta +(y-2)sin\theta=cos^2 \, \theta+sin^2\, \theta$
On compaining we get,

$(x-2) = cos\theta$ .....

$(i)$

$(y-2) = sin \theta$ .....

$(ii)$
On squaring and then adding Eqs.

$(i)$ and

$(ii),$ we get

$(x-2)^2+(y-2)^2=cos^2\theta +sin^2 \theta$

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

$\Rightarrow x^2+y^2-4x-4y+7=0$

The equation of the smallest circle passing through the points

$(2, 2)$ and

$(3, 3)$ is

0%
$x^{2} + y^{2} + 5x + 5y + 12 = 0$
0%
$x^{2} + y^{2} - 5x - 5y + 12 = 0$
0%
$x^{2} + y^{2} + 5x - 5y + 12 = 0$
0%
$x^{2} + y^{2} - 5x + 5y + 12 = 0$

Explanation

If the smallest circle is drawn through the points

$(2,2)$ and

$(3,3)$ then, the point have to be the opposite end of the diameter of the circle.
Therefore, the centre of the circle is

$C=(\dfrac{5}{2},\dfrac{5}{2})$ .
The radius of the circle is

$R=\dfrac{\sqrt{2}}{2}$ ....(using distance formula)
Hence equation of the circle is

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

$x^{2}+y^{2}-5x-5y+\dfrac{50}{4}=\dfrac{1}{2}$
Or

$x^{2}+y^{2}-5x-5y+\dfrac{25}{2}-\dfrac{1}{2}=0$
Or

$x^{2}+y^{2}-5x-5y+12=0$ .

Let

$ABCD$ be a square of side length

$1$ . and

$\Gamma$ a circle passing through

$B$ and

$C$ , and touching

$AD$ . The radius of

$\Gamma$ is

0%
$\cfrac { 3 }{ 8 }$
0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 1 }{ \sqrt { 2 } }$
0%
$\cfrac { 5 }{ 8 }$

Explanation

$PC=r$

${ PC }^{ 2 }={ r }^{ 2 }$

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

$1-2r+\cfrac { 1 }{ 4 } =0$

$r=\cfrac { 5 }{ 8 }$

If focii of

$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ coincide with the focii of

$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and eccentricity of the hyperbola is

$2$ , then

0%
$a^2+b^2=14$
0%
There is a director circle of the hyperbola
0%
Centre of the director circle is $(0,0)$
0%
Length of latus rectum of the hyperbola is $12$

Explanation

For the ellipse,

$a=5, b=3$ and

$e=\sqrt{\dfrac{25-9}{25}}=\dfrac{4}{5}$

$\therefore ae=4$
Hence, the focii are

$(-4, 0)$ and

$(4, 0)$

As the focii of ellipse and hyperbola coincide
For the hyperbola,

$ae=4, e=2$

$\therefore a=2$

$b^2=a^{2}(e^{2}-1)=4(4-1)=12$

$\therefore b=\sqrt{12}$

Length of latus rectum

$=\dfrac{2 b^2}{a}=2 \times \dfrac{12}{2}$

$=12$

Hence, option D is correct.

$e_{1}$ and

$e_{2}$ are the eccentricities of two conics with

$e_{1}^{2} + e_{2}^{2} = 3$ , then the conics are.

0%
Ellipses
0%
Parabolas
0%
Circles
0%
Hyperbolas

Explanation

Given,

${ e }_{ 1 }^{ 2 }+{ e }_{ 2 }^{ 2 }=3$

${ e }_{ 1 }$ and

${ e }_{ 2 }$ are eccentricities of same type of conic.

Let

${ e }_{ 1 }={ e }_{ 2 }$

$\therefore 2{ e }_{ 1 }^{ 2 }=3$ (from given equation)

${ e }_{ 1 }^{ 2 }=\cfrac { 3 }{ 2 } =1.5$

${ e }_{ 1 }=\sqrt { 1.5 } >1$

$\therefore { e }_{ 1 }={ e }_{ 2 }>1$

$\therefore$ Both conics are Hyperbolas.

$\because { e>1 }$

The line segment joining the foci of the hyperbola

$x^{2} - y^{2} + 1 = 0$ is one of the diameters of a circle. The equation of the circle is :

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

Explanation

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

Foci of the given hyperbola

$= (0, \pm \sqrt {2})$

Centre of the hyperbola is

$(0,0)$

Diameter of the circle is the distance between foci of the hyperbola and which is given by

$D=\sqrt{(0-0)^{2}+(\sqrt{2}+\sqrt{2})^{2}}=2\sqrt{2}$

Centre of the circle will be same as that of the hyperbola

$\therefore Centre = (0, 0)$ and

$Radius = \sqrt {2}$

$\therefore$ Equation of the circle is

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

The ends of the latus rectum of the parabola

$x^{2} + 10x - 16y + 25 = 0$ are

0%
$(3, 4), (-13, 4)$
0%
$(5, -8), (-5, 8)$
0%
$(3, -4), (13, 4)$
0%
$(-3, 4), (13, -4)$

Explanation

The equation

$x^{2} + 10x - 16y + 25 = 0$ can be written as

$(x+5)^2=16y$ ....

$(i)$ here

$a=4$ and vertex will be

$-5,0$ and focus will be

$-5,4$ .

So, the y-coordinate of ends of latus rectum will be

$4$ .

Putting in equation we have,

$(x+5)^2=16\times 4=64$

We have

$x=3,-13$

Substituting

$x$ in

$(i)$ , we get

$y=4,4$

Hence, ends of latus rectum will be

$(-3,4), (-13,4)$

The eccentricity of an ellipse

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

0%
$\dfrac { \sqrt { 3 } }{ 5 }$
0%
$\dfrac { \sqrt { 5 } }{ 3 }$
0%
$\dfrac { \sqrt { 7 } }{ 4 }$
0%
$\dfrac { 2 }{ 5 }$

Explanation

Given equation of an ellipse is

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

$\Rightarrow \dfrac { 9{ x }^{ 2 } }{ 144 } +\dfrac { 16{ y }^{ 2 } }{ 144 } =\dfrac { 144 }{ 144 }$

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

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

${ b }^{ 2 }=9$

$\left[ \because { a }^{ 2 } > { b }^{ 2 } \right]$

$\therefore$ Eccentricity

$\left( e \right) =\sqrt { 1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\dfrac { 9 }{ 16 } } =\sqrt { \dfrac { 7 }{ 16 } } =\dfrac { \sqrt { 7 } }{ 4 }$ .

The directrix of a parabola is

$x+8=0$ and its focus is at

$(4,3)$ . Then, the length of the latusrectum of the parabola is

0%
$5$
0%
$9$
0%
$10$
0%
$12$
0%
$24$

Explanation

Equation of parabola is

$\left| x+8 \right| =\sqrt { { (x-4) }^{ 2 }+{ (y-3) }^{ 2 } }$

$\Rightarrow {x}^{2}+64+16x={x}^{2}+16-8x+{(y-3)}^{2}$

$\Rightarrow { (y-3) }^{ 2 }=24(x+2)$

The foci of the ellipse

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

0%
$\left( \pm \cfrac { \sqrt { 3 } }{ 2 } ,0 \right)$
0%
$\left( \pm \cfrac { \sqrt { 5 } }{ 2 } ,0 \right)$
0%
$\left( \pm \cfrac { \sqrt { 5 } }{ 3 } ,0 \right)$
0%
$\left( \pm \cfrac { \sqrt { 5 } }{ 6 } ,0 \right)$
0%
$\left( \pm \cfrac { \sqrt { 5 } }{ 4 } ,0 \right)$

Explanation

Given,

$4{ x }^{ 2 }+9{ y }^{ 2 }=1..(i)$
We can write EQ.(i) as

$\quad \cfrac { { x }^{ 2 } }{ { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { \left( \cfrac { 1 }{ 3 } \right) }^{ 2 } } =1\quad$
Here,

$a=\cfrac { 1 }{ 2 } ;b=\cfrac { 1 }{ 3 }$

$\because$

$a> b$

$\because a>b$

$\therefore e=\sqrt { 1-\cfrac { \cfrac { 1 }{ 9 } }{ \cfrac { 1 }{ 4 } } } =\sqrt { 1-\cfrac { 4 }{ 9 } } =\cfrac { \sqrt { 5 } }{ 3 }$
Here foci

$=\left( \pm ae,0 \right)$

$=\left( \pm \cfrac { 1 }{ 2 } \cfrac { \sqrt { 5 } }{ 3 } ,0 \right) =\left( \pm \cfrac { \sqrt { 5 } }{ 6 } ,0 \right) \quad$

If the eccentricity of the hyperbola

${ x }^{ 2 }-{ y }^{ 2 }\sec ^{ 2 }{ \alpha } =5$ is

$\sqrt { 3 }$ times the eccentricity of the ellipse

${ x }^{ 2 }\sec ^{ 2 }{ \alpha } +{ y }^{ 2 }=25$ , then the value of

$\alpha$ is

0%
${ \pi }/{ 6 }$
0%
${ \pi }/{ 4 }$
0%
${ \pi }/{ 3 }$
0%
${ \pi }/{ 2 }$

Explanation

Equation for ellipse can be written as

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

the eccentricity of the given ellipse

$e_e^2=1-\dfrac{25\cos^2\alpha}{25}=\sin^2\alpha$

And the eccentricity of the Hyperbola is given as

$e_e^2=1+\dfrac{5\cos^2\alpha}{5}=1+\cos^2\alpha$

$e_h^2=3e_e^2$ (given in the question)

$1+\cos^2\alpha=3\sin^2\alpha\\ \sin^2\alpha=1/2\\ \alpha=\dfrac{\pi}{4}$

For the parabola

${ y }^{ 2 }+8x-12y+20=0$

0%
Vertex is $(2,6)$
0%
Focus is $(0.6)$
0%
Latusrectum $4$
0%
Axis $y=6$

Explanation

Given parabola can be rewritten as

${ (y-6) }^{ 2 }=-8(x-2)$
Which is of the form

${ Y }^{ 2 }=-4AX$
Where

$Y=y-6,X=x-2,A=2$

$\therefore$ Vertex

$(2.6)$
Focus

$(0.6)$
Latusrectum

$=4A=4\times 8$
Axis is

$y=6$

The foci of the conic section

$25{ x }^{ 2 }+16{ y }^{ 2 }-150x=175$ are

0%
$\left( 0,\pm 3 \right)$
0%
$\left( 0,\pm 2 \right)$
0%
$\left( 3,\pm 3 \right)$
0%
$\left( 0,\pm 1 \right)$

Explanation

$25{ x }^{ 2 }+16{ y }^{ 2 }-150x=175\\ { (5x) }^{ 2 }-150x+{ (15) }^{ 2 }+16{ y }^{ 2 }=175+{ (15) }^{ 2 }\\ { (5x-15) }^{ 2 }+16{ y }^{ 2 }=400\\ 25{ (x-3) }^{ 2 }+16{ y }^{ 2 }=400\\ \dfrac { { (x-3) }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$

Major axis of the ellipse is $y$ axis

So $a=5,b=4$

${ e }^{ 2 }=1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } =1-\dfrac { 16 }{ 25 } =\dfrac { 9 }{ 25 } \\ \Rightarrow e=\dfrac { 3 }{ 5 }$

Coordinates of foci of standard ellipse with major axis as $y$ are $(0,\pm ae)$

$\Rightarrow$ foci of given ellipse is

$\left( 3,\pm 5\times \dfrac { 3 }{ 5 } \right) \\ \left( 3,\pm 3 \right)$ '

So opttion $C$ is correct

The equation of directrix of the parabola

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

0%
$x = -1$
0%
$x = 1$
0%
$x = -\dfrac {3}{2}$
0%
$x = \dfrac {3}{2}$

Explanation

We have,

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

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

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

$y + 2 = Y, x - \dfrac {1}{2} = X$
We have,

$Y^{2} = -4X$
This is a parabola with directrix at

$X = 1$

$\Rightarrow x - \dfrac {1}{2} = 1$

$\Rightarrow x = \dfrac {3}{2}$ .

The radius of the circle passing through the points

$(2,3),(2,7)$ and

$(5,3)$ is

0%
$5$
0%
$4$
0%
$\cfrac { 5 }{ 2 }$
0%
$2$
0%
$\sqrt { 5 }$

Explanation

Let the given points

$A(2,3),B(5,3)$ and

$C(2,7)$
Diameters of circle is distance between

$B(5,3)$ and

$C(2,7)$

$d=\sqrt { { (2-5) }^{ 2 }+{ (7-3) }^{ 2 } } =\sqrt { 9+16 }$

$d=\sqrt { 25 } =5$

$\therefore$ Radius

$=\cfrac { d }{ 2 } =\cfrac { 5 }{ 2 }$

The one end of the latusrectum of the parabola

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

0%
$\left( 0,-1 \right)$
0%
$\left( 0,1 \right)$
0%
$\left( 0,-3 \right)$
0%
$\left( 3,0 \right)$
0%
$\left( 0,2 \right)$

Explanation

Given equation is

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

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

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

$\Rightarrow { \left( y-1 \right) }^{ 2 }=4\left( x+1 \right)$
Shift the origin to

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

$\Rightarrow { Y }^{ 2 }=4X$
Here focus is at

$\left( 1,0 \right)$ .
Hence, focus of original parabola becomes

$\left( 1-1,0+1 \right) =\left( 0,1 \right)$
Therefore, equation of latusrectum is

$x=0$ .
Thus point of intersection of parabola and latus rectum is

${ y }^{ 2 }-2y-3=0\Rightarrow y=-1$ or

$3$
So, the required points are

$\left( 0,-1 \right) ,\left( 0,3 \right)$ .

The lines

$y=x+\sqrt { 2 }$ and

$y=x-2\sqrt { 2 }$ are the tangent of certain circle. If the point

$\left( 0,\sqrt { 2 } \right)$ lies on this circle, then its equation is

0%
${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 }$
0%
${ \left( x+\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 }$
0%
${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y-\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 }$
0%
None of the above

Explanation

Both tangents are parallel to each other, Therefore the distance between these two lines is the diameter of the circle.

diameter

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

Radius is =

$\dfrac{3}{2}$

$(x_1, y_1)$ is the centre of the circle, Then

$\dfrac{x_1-0}{1}=\dfrac{y_1-\sqrt{2}}{-1}=\dfrac{1}{2}\dfrac{|0-\sqrt{2}-2\sqrt{2}|}{2}$

Centre is

$\bigl(\dfrac{3}{2\sqrt{2}}, \dfrac{1}{2\sqrt{2}}\Bigr)$

Equation of the circle is

$(x-\frac{1}{3\sqrt{2}})^2+(y-\frac{1}{2\sqrt{2}})^2=9/4$

On the parabola

$y={ x }^{ 2 }$ , the point least distant from the straight line

$y=2x-4$ is

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

Explanation

Given, parabola is

$y={ x }^{ 2 }$ ....(i)
and straight line is

$y=2x-4$ ....(ii)
From equations (i) and (ii), we get

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

$f\left( x \right) ={ x }^{ 2 }-2x-4$
Thus

$f^{ ' }\left( x \right) =2x-2$
For least distance, put

$f^{ ' }\left( x \right) =0$

$\Rightarrow 2x-2=0$

$\Rightarrow x=1$
From equation (i), we have

$y=1$
Hence, the point least distant from the line is

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

If the eccentricity of the ellipse

$a{ x }^{ 2 }+4{ y }^{ 2 }=4a,(a<4)$ is

$\cfrac { 1 }{ \sqrt { 2 } }$ , then its semi-minor axis is equal to

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

Explanation

Given equation of ellipse is

$a{ x }^{ 2 }+4{ y }^{ 2 }=4a...(i)\quad$
We can write Eq.(i) as

$\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ a } =1$
Here

$\quad { a }^{ 2 }=4;{ b }^{ 2 }=a$

$\because a<4$

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

$\Rightarrow \cfrac { 1 }{ \sqrt { 2 } } =\sqrt { 1-\cfrac { a }{ 4 } } \Rightarrow \cfrac { 1 }{ 2 } =1-\cfrac { a }{ 4 }$

$\Rightarrow \cfrac { a }{ 4 } =\cfrac { 1 }{ 2 } \Rightarrow a=2$

$\therefore { b }^{ 2 }=2\Rightarrow b=\sqrt { 2 }$
Hence, semi-minor axis is

$b=\sqrt{2}$

The lines

$2x-3y=5$ and

$3x-4y=7$ are the diameters of a circle of area

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

0%
${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=62$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=47\quad$
0%
${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=47$
0%
${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=62\quad$

Explanation

Given equation of lines is

$2x-3y=5$ ..... (i) and

$3x-4y=7$ .....(ii)

Solving above equations, we get the point of intersection as

$(1,-1)$ .

Eqns (i) and (ii) are diameters of the circle.

We know that the centre of circle

$=$ point of intersection of diameters

$=(1,-1)$ .

Now, it is given that the area of the circle

$=154$ .

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

$\Rightarrow r=7$
Hence, the equation of required circle is

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

$\Rightarrow$

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

The equation of the circle which touches the lines

$x = 0, y = 0$ and

$4x + 3y = 12$ is

0%
$x^{2} + y^{2} - 2x - 2y - 1 = 0$
0%
$x^{2} + y^{2} - 2x - 2y + 3 = 0$
0%
$x^{2} + y^{2} - 2x - 2y + 2 = 0$
0%
$x^{2} + y^{2} - 2x - 2y + 1 = 0$
0%
$x^{2} + y^{2} - 2x - 2y - 3 = 0$

Explanation

Let the radius of the circle be

$r$ .

Since, circle touches the lines

$x = 0, y = 0$ , i.e.,

$x$ -axis and

$y$ -axis, then its centre is

$C\equiv (r, r)$

Now,

$\text{radius} (PC) =$ Perpendicular distance from

$(C)$ to the line

$4x + 3y = 12$

i.e.,

$r = \dfrac {|4r + 3r - 12|}{\sqrt {16 + 9}}$

$\Rightarrow 7r - 12 = \pm 5r$

So,

$r = 1$ or

$6$

But

$r\neq 6$

Required equation of circle is

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

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

$\Rightarrow x^{2} + y^{2} - 2x - 2y + 1 = 0$

The eccentricity of the ellipse

$12x^2 + 7y^2 = 84$ is equal to :

0%
$\dfrac {\sqrt5}{7}$
0%
$\sqrt{\dfrac {5}{7}}$
0%
$\dfrac {\sqrt5}{12}$
0%
$\dfrac {5}{7}$
0%
$\dfrac {7}{12}$

Explanation

Given equation is

$12x^2+7y^2=84$

It can be rewritten as

$\dfrac {x^2}{84/12} + \dfrac {y^2}{84/7} = 1$

$\Rightarrow \dfrac {x^2}{7} + \dfrac {y^2}{12} = 1$
Here,

$a^2 = 7 , b^2 = 12$ i.e.,

$b>a$
Therefore,

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

$= \sqrt{1 - \dfrac{7}{12}}$

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

Equations

$x = a\cos \theta$ and

$y= b\sin \theta$ represent a conic section whose eccentricity

$e$ is given by

0%
$e^{2} = \dfrac {a^{2} + b^{2}}{a^{2}}$
0%
$e^{2} = \dfrac {a^{2} + b^{2}}{b^{2}}$
0%
$e^{2} = \dfrac {a^{2} - b^{2}}{a^{2}}$
0%
None of these

Explanation

Given equation can be rewritten as

$\cos \theta = \dfrac {x}{a}$ and

$\sin \theta = \dfrac {y}{b}$

$\because \cos^{2}\theta +\sin^{2} \theta = 1$

$\therefore \left (\dfrac {x}{a}\right )^{2} + \left (\dfrac {y}{b}\right )^{2} = 1$
Thus, it represents an equation of an ellipse.

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

An ellipse has

$OB$ as semi-minor axis,

$F$ and

${ F }^{ ' }$ its foci and the

$\angle FB{ F }^{ ' }$ is a right angle. Then, the eccentricity of the ellipse is

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

Explanation

Given,

$F$ and

${ F }^{ ' }$ are foci of an ellipse, whose coordinates are

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

$\left( -ae,0 \right)$ respectively and coordinates of

$B$ are

$\left( 0,b \right)$ .
Slope of

$BF=\dfrac { b }{ -ae }$
and slope of

$B{ F }^{ ' }=\dfrac { b }{ ae }$
Since,

$\angle FB{ F }^{ ' }={ 90 }^{ o }$
Thus

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

$\Rightarrow { b }^{ 2 }={ a }^{ 2 }{ e }^{ 2 }$
Here

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

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

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

The hyperbola

$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ passes through the point

$\left( \sqrt { 6 } ,3 \right)$ and the length of the latusrectum is

$\cfrac { 18 }{ 5 }$ . Then, the length of the transverse axis is equal to

0%
$5$
0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

Given, equation of hyperbola

$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1...(i)\quad$
Eq. (i) passes through

$\left( \sqrt { 6 } ,3 \right)$

$\therefore \cfrac { { (\sqrt { 6 } ) }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { 3 }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow \cfrac { 6 }{ { a }^{ 2 } } -\cfrac { 9 }{ { b }^{ 2 } } =1...(ii)$

$\cfrac { 2{ b }^{ 2 } }{ a } =\cfrac { 18 }{ 5 } ....(iii)$
On solving Eqs. (i) and (ii)

${ a }^{ 2 }=1,{ b }^{ 2 }=\cfrac { 9 }{ 5 }$

The equation of the circumcircle of the triangle formed by the lines

$y + \sqrt{3} x = 6, y - \sqrt{3} x = 6$ and

$y=0$ is

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

Explanation

Triangle

$ABC$ is an equilateral triangle

Hence,

$BC=4\sqrt{3} \\ 2R=\cfrac{a}{\sin A} \\ 2R=\cfrac{4\sqrt{3}}{\sin 60^o}=8 \\ R=4$

Circumcentre of

$ABC$

$\left( \cfrac{x_1+x_2+x_3}{3},\cfrac{y_1+y_2+y_3}{3} \right) \\ =(0,2)$

Equation of circumcircle

$(x-0)^2+(y-2)^2=4^2 \\ x^2+y^2-4y=12$

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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