Conic Sections Test

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Conic Sections Test - 54

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Weekly Quiz Competition

Question 1
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$$S_{1}$$ and $$S_{2}$$ are the foci of an ellipse of major axis of length 10 units, and P is any point on the ellipse such that the perimeter or triangle $$PS_{1}S_{2}$$ is 15. Then the eccentricity of the ellipse is

A
0.5

B
0.25

C
0.28

D
0.75

Solution
Question 2
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$$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad $$

A
$$(1,1)$$

B
$$(0,2)$$

C
$$(2,0)$$

D
$$None\quad of\quad these$$

Solution

Given Hyperbola: $$x^{2}+4y^{2}+6xy+8x-2y+7=0$$
Centre: Point of intersection of asymptotes of hyperbola.
Now finding Asymptotes of given equation, taking equation of asymptote $$y=mx+c$$ by replacing $$x\rightarrow 1, y\rightarrow m$$ in $$\phi_n(m)$$
When $$n=2$$, $$\phi_{2}(m)=1+4m^{2}+6m$$
$$\phi_{1}(m)=8-2m$$
$$\phi_{0}(m)=7$$
$$\phi_{2}^{1}(m)=8m+6$$
$$\phi_{1}(m)=8-2m$$

Putting $$\phi_{2}(m)=0$$
$$\Rightarrow 1+6m+4m^{2}=0$$
$$\Rightarrow m=\cfrac{-6\pm \sqrt {36-16}}{2(4)}$$
$$\Rightarrow m=\cfrac{-6\pm\sqrt 20}{2(4)}$$
$$\Rightarrow m=\cfrac{-3\pm \sqrt 5}{4}$$
So, m$$=\cfrac{-3+\sqrt 5}{4}, \cfrac{-3-\sqrt 5}{4}$$
Value of $$c$$, when $$m$$ is different
$$c= \cfrac{-\phi_{1}(m)}{\phi_{2}^{'}(m)}=\cfrac{8-2m}{8m+6}$$

For $$m=\cfrac{-3+\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3+\sqrt 5}{4})}{8\cfrac{-3+\sqrt 5}{4})+6}=\cfrac{3-\sqrt 5+16}{-6+2\sqrt 5+6}=\cfrac{19\sqrt 5-5}{10}$$

For $$m=\cfrac{-3-\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3-\sqrt 5}{4})}{8\cfrac{-3-\sqrt 5}{4})+6}=\cfrac{19+15}{-2 \sqrt 5}=\cfrac{-(19\sqrt 5+5)}{10}$$
Equation of Asymptotes : $$y=\cfrac{-3+\sqrt 5}{4}x +\cfrac{19\sqrt 5-5}{10}$$ & $$y=\cfrac{-3-\sqrt 5}{4}x-(\cfrac{5+19\sqrt 5}{10})$$
On solving them for x & y, putting LHS-RHS
$$\Rightarrow 0=\cfrac{\sqrt 5}{2}x+\cfrac{19\sqrt 5}{5} \Rightarrow x=\cfrac{-38}{5}$$
Now putting values of x in Asymptotes equation, we get
$$y=\cfrac{-3-\sqrt 5(-19)}{10}+\cfrac{+5+19\sqrt 5}{10}=\cfrac{+52}{10}$$
Centre$$(\cfrac{-38}{5}, \cfrac{+52}{10})$$.
Question 3
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'O' is the vertex of the parabola $${ y }^{ 2 }=8x$$ and L is the upper end of the latus rectum. If LH is drawn perpendicular to OL meeting OX in H, then the length of the double ordinate through H is $$\lambda \sqrt { 5 } $$ where $$\lambda $$ is equal to

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution
Question 4
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Which of the following equations does not represent a hyperbola?

A
$$xy = 4$$

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

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

D
$$
x^2 - 4xy + 3y^2 = 1
$$

Solution
Question 5
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The set of points $$(x, y)$$ whose distance from the line $$y = 2x + 2$$ is the same as the distance from $$(2, 0)$$ is a parabola. This parabola is congruent to the parabola in standard form $$y = Kx^{2}$$ for some $$K$$ which is equal to

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

B
$$\dfrac {\sqrt {5}}{4}$$

C
$$\dfrac {4}{\sqrt {5}}$$

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

Solution

Distance from $$(2, 0)$$ to $$2x - y + 2 = 0$$ is equal to $$\dfrac {6}{\sqrt {5}} =$$ semi latus rectum
$$\therefore$$ length of latus rectum is $$\dfrac {12}{\sqrt {5}}$$
$$\therefore x^{2} = \dfrac {1}{K}y \Rightarrow \dfrac {1}{K} = \dfrac {12}{\sqrt {5}}$$
$$\Rightarrow K = \dfrac {\sqrt {5}}{12}$$.
Question 6
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Directions For Questions

Circle touching a line $$L=0$$ at a point $$\left({x}_{1},{y}_{1}\right)$$ on it is
$${\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}+\lambda L=0,\lambda\in R$$
Circle through the two points $$A\left({x}_{1},{y}_{1}\right)$$ and $$B\left({x}_{2},{y}_{2}\right)$$ is
$$\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)+\left(y-{y}_{1}\right)\left(y-{y}_{2}\right)+\lambda L=0 , \lambda \in R$$
where $$L=0$$ is the equation of the line $$AB$$
On the basis of the above information,answer the following questions:

...view full instructions

From the point $$A\left(0,3\right)$$ on the circle $${x}^{2}-4x+{\left(y-3\right)}^{2}=0$$ a chord $$AB$$ is drawn and extended to a point $$M$$ such that $$AM=2AB$$.The locus is

A
$${x}^{2}+{y}^{2}-8x-6y+9=0$$

B
$${x}^{2}+{y}^{2}+8x-6y+9=0$$

C
$${x}^{2}+{y}^{2}-8x+6y+9=0$$

D
$${x}^{2}+{y}^{2}+8x+6y+9=0$$

Solution

$${x}^{2}-4x+{\left(y-3\right)}^{2}=0$$
Add $$4$$ both sides of the above equation, we get
$$\Rightarrow {x}^{2}-4x+4+{\left(y-3\right)}^{2}=4$$
$$\Rightarrow {\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=4$$
Take $$B=\left(2+2\cos\theta,3+2\sin\theta\right)$$
$$\therefore 2+2\cos\theta=\dfrac{x+0}{2} , 3+2\sin\theta=\dfrac{y+3}{2}$$
or $$2\cos\theta=\dfrac{x}{2}-2, 2\sin\theta=\dfrac{y+3}{2}-3$$
or $$2\cos\theta=\dfrac{x-4}{2}, 2\sin\theta=\dfrac{y-3}{2}$$
Eliminating $$\theta$$ we get
$${\left(2\cos\theta\right)}^{2}+{\left(2\sin\theta\right)}^{2}={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$$
$$4={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$$
$${\left(x-4\right)}^{2}+{\left(y-3\right)}^{2}=16$$
On simplifying , we get
$${x}^{2}+{y}^{2}-8x-6y+9=0$$
Question 7
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The line $$2x + y = 1$$ is tangent to the hyperbola $$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. If this line passes through the point of intersection of the nearest directrix and the x-axis, then eccentricity of the hyperbola.

A
1

B
2

C
3

D
4

Solution
Question 8
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The equation of the circle having the lines $$x^2 + 2xy + 3x + 6y = 0$$ as its normals and having size just sufficient to contain the circle $$x (x - 4) + y(y - 3) = 0$$ is

A
$$x^2 + y^2 + 3x - 6y - 40 = 0$$

B
$$x^2 + y^2 + 6x - 3y - 45 = 0$$

C
$$x^2 + y^2 + 8x + 4y - 20 = 0$$

D
$$x^2 + y^2 + 4x + 8y + 20 =0$$

Solution

Pair of lines:
$$x^{2}+2xy+3x+6y=0$$
$$\Rightarrow x(x+2y)+3(x+2y)=0$$
$$\Rightarrow (x+2y)(x+3)=0$$
$$\Rightarrow x+2y=0$$
and $$x+3=0$$
Are the two normals and their point of intersection must be the centre of the circle.
$$\Rightarrow x=-3$$
and $$y=\dfrac{-x}{2}=\dfrac{3}{2}$$
$$\Rightarrow (-3,\dfrac{3}{2}) $$ is the centre of required circle.
($$\because x(x-4)+y(y-3)=0$$ is the diametric form of the circle. Hence, $$(0,0)$$ and $$(4,3)$$ are the diametric end points.)
$$\Rightarrow$$ Centre $$\rightarrow (\dfrac{0+4}{2},\dfrac{0+3}{2})\rightarrow (2,\dfrac{3}{2})$$
radius $$=\dfrac{1}{2}(\sqrt{16+9})$$
$$=\dfrac{5}{2}$$
The required circle with centre $$(-3,\dfrac{3}{2})$$ is just sufficient to contain the circle
$$x(x-4)+y(y-3)=0$$
$$\therefore$$ radius of required circle
=distance between $$(-3,\dfrac{3}{2})$$ and $$(2,\dfrac{3}{2})$$ $$+$$ radius of given circle.
$$=\sqrt{(-3-2)^{2}+0}+\dfrac{5}{2}$$
$$=5+\dfrac{5}{2}=\dfrac{15}{2}$$
$$\therefore$$ Equation of required circle:
$$\Rightarrow (x+3)^{2}+(y-\dfrac{3}{2})^{2}=\dfrac{225}{4}$$
$$\Rightarrow x^{2}+y^{2}+9+\dfrac{9}{4}+6x-3y=\dfrac{225}{4}$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9+\dfrac{9}{4}-\dfrac{225}{4}=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9-\dfrac{216}{4}=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9-54=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y-45=0$$
$$\therefore B)$$ Answer.
Question 9
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$$LL^1$$ is the latus rectum of an ellipse and $$\Delta S^1LL^1$$ is an equilateral triangle. Then $$e=?$$

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

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

C
$$\dfrac{1}{\sqrt{5}}$$

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

Solution

Let equation of the ellipse is $$\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ b^{ 2 } } =1$$

Refer the figure. By distance formula,
$${ S }^{ ' }L=\sqrt { { \left( ae-\left( -ae \right) \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$$
$$\therefore { S }^{ ' }L=\sqrt { { \left( 2ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } \right) }^{ 2 } }$$
$$\therefore { S }^{ ' }L=\sqrt { 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } }$$
$$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }=4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } }$$ (1)

Similarly, $$L{ L }^{ ' }=\sqrt { { \left( ae-ae \right) }^{ 2 }+{ \left( \frac { { b }^{ 2 } }{ a } -\left( -\frac { { b }^{ 2 } }{ a } \right) \right) }^{ 2 } }$$
$$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } \right) }^{ 2 } } $$
$$\therefore L{ L }^{ ' }=\sqrt { { \left( \frac { 2{ b }^{ 2 } }{ a } \right) }^{ 2 } } $$
$$\therefore L{ L }^{ ' }=\sqrt { { \frac { 4{ b }^{ 4 } }{ { a }^{ 2 } } } } $$
$$\therefore { \left( L{ L }^{ ' } \right) }^{ 2 }=\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } } $$ (2)

Now, given $$\triangle { S }^{ ' }LL^{ ' }$$ is equilateral triangle
$$\therefore { \left( { S }^{ ' }L \right) }^{ 2 }={ \left( L{ L }^{ ' } \right) }^{ 2 }$$

$$\therefore 4{ a }^{ 2 }{ e }^{ 2 }+\frac { { b }^{ 4 } }{ { a }^{ 2 } } =\frac { 4{ b }^{ 4 } }{ { a }^{ 2 } }$$

$$\therefore 4{ a }^{ 2 }{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 2 } } $$

$$\therefore 4{ e }^{ 2 }=\frac { 3{ b }^{ 4 } }{ { a }^{ 4 } } $$ (3)

Now, $$\frac { { b }^{ 2 } }{ { a }^{ 2 } } =1-{ e }^{ 2 }$$
$$\therefore { \left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) }^{ 2 }={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$
$$\therefore \frac { { b }^{ 4 } }{ { a }^{ 4 } } ={ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$ (4)

From equations (3) and (4), we can write,
$$4{ e }^{ 2 }=3{ \left( 1-{ e }^{ 2 } \right) }^{ 2 }$$

$$\therefore 4{ e }^{ 2 }=3\left( 1-2{ e }^{ 2 }+{ e }^{ 4 } \right)$$

$$\therefore 4{ e }^{ 2 }=3-6{ e }^{ 2 }+3{ e }^{ 4 }$$

$$\therefore 3{ e }^{ 4 }-10{ e }^{ 2 }+3=0$$

$$\therefore 3{ e }^{ 4 }-9{ e }^{ 2 }-{ e }^{ 2 }+3=0$$

$$\therefore 3{ e }^{ 2 }\left( { e }^{ 2 }-3 \right) -1\left( { e }^{ 2 }-3 \right) =0$$

$$\therefore \left( 3{ e }^{ 2 }-1 \right) \left( { e }^{ 2 }-3 \right) =0$$

$$\therefore { e }^{ 2 }=\frac { 1 }{ 3 } $$ or $${ e }^{ 2 }=3$$

$$\therefore { e }=\frac { 1 }{ \sqrt { 3 } } $$ or $${ e }=\sqrt { 3 }$$

But $${ e }<1$$
$$\therefore { e }\neq \sqrt { 3 } $$

$$\therefore { e }=\frac { 1 }{ \sqrt { 3 } } $$
Question 10
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The normal at $$ P(8, 8)$$ to the parabola $$y^2=8x$$ cuts it again at Q then PQ =

A
$$10$$

B
$$10\sqrt5$$

C
$$4\sqrt5$$

D
$$50$$

Solution