Conic Sections Test

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Conic Sections Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The ends of the latus rectum of the parabola $$x^{2} + 10x - 16y + 25 = 0$$ are

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

B
$$(5, -8), (-5, 8)$$

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

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

Solution

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 $$(-3,4), (-13,4)$$
Question 2
1 / -0

If $$e_{1}$$ and $$e_{2}$$ are the eccentricities of two conics with $$e_{1}^{2} + e_{2}^{2} = 3$$, then the conics are.

A
Ellipses

B
Parabolas

C
Circles

D
Hyperbolas

Solution

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. $$\because { e>1 } $$
Question 3
1 / -0

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 :

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

B
$$x^{2} + y^{2} = \sqrt {2}$$

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

D
$$x^{2} + y^{2} = 2\sqrt {2}$$

Solution

$$x^{2} - y^{2} + 1 = 0\implies (x-0)^{2}+(y-0)^{2}=0$$
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 $$x^{2} + y^{2} = 2$$
Question 4
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Consider the conic $$e{ x }^{ 2 }+\pi { y }^{ 2 }-2{ e }^{ 2 }x-2{ \pi }^{ 2 }y+{ e }^{ 3 }+{ \pi }^{ 3 }=\pi e$$. Suppose $$P$$ is any point on the conic and $${ S }_{ 1 }, { S }_{ 2 }$$ are the foci of the conic, then the maximum value of $$\left( P{ S }_{ 1 }+P{ S }_{ 2 } \right) $$ is

A
$$\pi e$$

B
$$\sqrt { \pi e } $$

C
$$2\sqrt { \pi } $$

D
$$2\sqrt { e } $$

Solution

$$\displaystyle{e{ x }^{ 2 }+\pi { y }^{ 2 }+-2{ e }^{ 2 }x-2{ \pi }^{ 2 }y+{ e }^{ 3 }+{ \pi }^{ 3 }=\pi e\\ e({ x }^{ 2 }-2ex+{ e }^{ 2 })+\pi ({ y }^{ 2 }-2\pi y+{ \pi }^{ 2 })=\pi e\\ \quad \dfrac { { \left( x-e \right) }^{ 2 } }{ \pi } +\dfrac { { \left( y-\pi \right) }^{ 2 } }{ e } =1\\ }$$
$$a=\sqrt { \pi } ,b=\sqrt { e } $$
for ellipse sum of focal distance $$PS+PS^{\prime} $$ is $$2a$$
So $$PS+PS^{\prime} =2\sqrt { \pi } $$
option $$C$$ is correct
Question 5
1 / -0

The eccentricity of an ellipse $$9{ x }^{ 2 }+16{ y }^{ 2 }=144$$ is

A
$$\dfrac { \sqrt { 3 } }{ 5 } $$

B
$$\dfrac { \sqrt { 5 } }{ 3 } $$

C
$$\dfrac { \sqrt { 7 } }{ 4 } $$

D
$$\dfrac { 2 }{ 5 } $$

Solution

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 $$ \left( e \right) =\sqrt { 1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } } =\sqrt { 1-\dfrac { 9 }{ 16 } } =\sqrt { \dfrac { 7 }{ 16 } } =\dfrac { \sqrt { 7 } }{ 4 } $$.
Question 6
1 / -0

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

A
$$5$$

B
$$9$$

C
$$10$$

D
$$12$$

E
$$24$$

Solution

Equation of parabola is
$$\left| x+8 \right| =\sqrt { { (x-4) }^{ 2 }+{ (y-3) }^{ 2 } } $$
$$\Rightarrow {x}^{2}+64+16x={x}^{2}+16-8x+{(y-3)}^{2}$$
$$\Rightarrow { (y-3) }^{ 2 }=24(x+2)$$
Question 7
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The one end of the latusrectum of the parabola $${ y }^{ 2 }-4x-2y-3=0$$ is at

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

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

C
$$\left( 0,-3 \right) $$

D
$$\left( 3,0 \right) $$

E
$$\left( 0,2 \right) $$

Solution

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 $$\left( 0,-1 \right) ,\left( 0,3 \right) $$.
Question 8
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The foci of the ellipse $$4{x}^{2}+9{y}^{2}=1$$ are

A
$$\left( \pm \cfrac { \sqrt { 3 } }{ 2 } ,0 \right) $$

B
$$\left( \pm \cfrac { \sqrt { 5 } }{ 2 } ,0 \right) $$

C
$$\left( \pm \cfrac { \sqrt { 5 } }{ 3 } ,0 \right) $$

D
$$\left( \pm \cfrac { \sqrt { 5 } }{ 6 } ,0 \right) $$

E
$$\left( \pm \cfrac { \sqrt { 5 } }{ 4 } ,0 \right) $$

Solution

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$$=\left( \pm ae,0 \right) $$
$$=\left( \pm \cfrac { 1 }{ 2 } \cfrac { \sqrt { 5 } }{ 3 } ,0 \right) =\left( \pm \cfrac { \sqrt { 5 } }{ 6 } ,0 \right) \quad $$
Question 9
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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

A
$${ \pi }/{ 6 }$$

B
$${ \pi }/{ 4 }$$

C
$${ \pi }/{ 3 }$$

D
$${ \pi }/{ 2 }$$

Solution

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)
$$1+\cos^2\alpha=3\sin^2\alpha\\ \sin^2\alpha=1/2\\ \alpha=\dfrac{\pi}{4}$$
Question 10
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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

A
$$2$$

B
$$\sqrt { 2 } $$

C
$$1$$

D
$$\sqrt { 3 } $$

E
$$3$$

Solution

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 $$b=\sqrt{2}$$

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Conic Sections Test - 26

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Conic Sections Test - 26

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Question 1

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

$$(5, -8), (-5, 8)$$

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

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

Solution

Question 2

Ellipses

Parabolas

Circles

Hyperbolas

Solution

Question 3

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

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

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

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

Solution

Question 4

$$\pi e$$

$$\sqrt { \pi e } $$

$$2\sqrt { \pi } $$

$$2\sqrt { e } $$

Solution

Question 5

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

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

$$\dfrac { \sqrt { 7 } }{ 4 } $$

$$\dfrac { 2 }{ 5 } $$

Solution

Question 6

$$5$$

$$9$$

$$10$$

$$12$$

$$24$$

Solution

Question 7

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

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

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

$$\left( 3,0 \right) $$

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

Solution

Question 8

$$\left( \pm \cfrac { \sqrt { 3 } }{ 2 } ,0 \right) $$

$$\left( \pm \cfrac { \sqrt { 5 } }{ 2 } ,0 \right) $$

$$\left( \pm \cfrac { \sqrt { 5 } }{ 3 } ,0 \right) $$

$$\left( \pm \cfrac { \sqrt { 5 } }{ 6 } ,0 \right) $$

$$\left( \pm \cfrac { \sqrt { 5 } }{ 4 } ,0 \right) $$

Solution

Question 9

$${ \pi }/{ 6 }$$

$${ \pi }/{ 4 }$$

$${ \pi }/{ 3 }$$

$${ \pi }/{ 2 }$$

Solution

Question 10

$$2$$

$$\sqrt { 2 } $$

$$1$$

$$\sqrt { 3 } $$

$$3$$

Solution

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