Conic Sections Test

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Conic Sections Test - 16

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Question 1
1 / -0

Length of the latusrectum of the hyperbola $$xy=c$$ ,is equal to

A
$$2c$$

B
$$\sqrt{2}c$$

C
$$2\sqrt{2}c$$

D
$$4c$$

Solution

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 $$2\dfrac{b^2}{a}=\dfrac{4c}{\sqrt{2c}}=2\sqrt2 c$$
Question 2
1 / -0

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

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

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

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

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

Solution

$$F= \left ( \dfrac{u^{2}}{2g}\sin \ 2\alpha ,\dfrac{-u^{2}}{2g}\cos \ 2\alpha \right )$$
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.
Question 3
1 / -0

The difference between the length $$2a$$ of the transverse axis of a hyperbola of eccentricity $$e$$ and the length of its latus rectum is :

A
$$2a\left| 3-{ e }^{ 2 } \right| $$

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

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

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

Solution

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 $$=2a\left| 2-{ e }^{ 2 } \right| $$
Question 4
1 / -0

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

A
$$y^{2}=4a(x+a)$$

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

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

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

Solution

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,
$$PS=PM$$
$$(x+2a)^2+y^2=x^2$$
$$\Rightarrow y^2=-4ax-4a^2$$
$$\Rightarrow y^2=-4a(x+a)$$
Question 5
1 / -0

If $$L_{1}L_{2}$$ is the latusrectum of $$y^{2}=12x$$, $$P$$ is any point on the directrix then the area of $$\Delta PL_{1}L_{2}$$ =

A
$$32$$

B
$$18$$

C
$$36$$

D
$$16$$

Solution

The length of latus rectum is $$2l=4a=12$$
and distance between latus rectum and directrix is $$2a=6$$
Therefore, $$\triangle PL_1L_2=\dfrac{(2a)(2l)}{2}=36 $$
Question 6
1 / -0

The length of latus rectum of the parabola $$(x-2a)^{2}+y^{2}=x^{2}$$ is

A
$$2a$$

B
$$3a$$

C
$$6a$$

D
$$4a$$

Solution

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 $$=4\times a = 4a$$
Question 7
1 / -0

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

A
$$B,D,C,A$$

B
$$A,B,C,D$$

C
$$A,C,D,B.$$

D
$$A,D,C,B$$

Solution

We find the latus rectum for the given parabolas.
A. $$y=4{ x }^{ 2 }+x+1$$ $$\Rightarrow y-1=4{ x }^{ 2 }+x$$
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 $$\displaystyle\frac{1}{4}$$

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

C. $$x={ 2y }^{ 2 }+y+3$$
$$\Rightarrow x-3={ 2y }^{ 2 }+y$$
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 $$\displaystyle\frac{1}{2}$$

D. $${ y }^{ 2 }+y+x+9=0$$ $$\Rightarrow x-9={ y }^{ 2 }+y$$
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 $$A,C,D,B$$
Question 8
1 / -0

The length of latus rectum of the parabola $$y^{2}+8x-2y+17=0$$ is

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution

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 $$=\displaystyle 4a = 4\times 2=8$$
Question 9
1 / -0

The length of latus rectum of the hyperbola $$xy-3x-3y+7=0$$ is

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

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 $$=2 \times 2=4$$
Question 10
1 / -0

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

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

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

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

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

Solution

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.

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Conic Sections Test - 16

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Conic Sections Test - 16

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Question 1

$$2c$$

$$\sqrt{2}c$$

$$2\sqrt{2}c$$

$$4c$$

Solution

Question 2

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

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

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

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

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

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

Solution

Question 5

$$32$$

$$18$$

$$36$$

$$16$$

Solution

Question 6

$$2a$$

$$3a$$

$$6a$$

$$4a$$

Solution

Question 7

$$B,D,C,A$$

$$A,B,C,D$$

$$A,C,D,B.$$

$$A,D,C,B$$

Solution

Question 8

$$2$$

$$4$$

$$8$$

$$16$$

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

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

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

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

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

Solution

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