Conic Sections Test

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Conic Sections Test - 31

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Weekly Quiz Competition

Question 1
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Equation of the parabola having focus $$(3,2)$$ and Vertex $$(-1,2)$$ is

A
$$(x+1)^2=16(y-2)$$

B
$$(x-1)^2=16(y+2)$$

C
$$(y-2)^2=16(x+1)$$

D
$$(y+2)^2=16(x-1)$$

Solution

Equation of a General Parabola having axis along the x-axis is given by $$y^2=4ax$$

where $$a=$$ distance between Vertex and Focus.

Here $$y$$ coordinate in Focus and Vertex is the same.

Thus this parabola would be along direction of the x-axis.

Diatance between vertex and focus $$=\sqrt{(3-(-1))^2+(2-2)^2}=4$$

thus equation of parabola is $$(y-2)^2=4.(4)(x-(-1))$$

i.e.$$(y-2)^2=16(x+1)$$
Question 2
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Directions For Questions

Equation of a circle is $$S={x}^{2}+{y}^{2}+2gx+2fy+c$$
Its notation is $${S}_{1}={x}_{1}x+{y}_{1}y+g\left(x+{x}_{1}\right)+f\left({y}_{1}+y\right)+c$$
$${S}_{11}={x}_{1}^{2}+{y}_{1}^{2}+2g{x}_{1}+2f{y}_{1}+c$$
$$\left(i\right)$$Location of $$P\left({x}_{1},{y}_{1}\right):$$
$$P$$ lies inside the circle $$S=0$$, if $${S}_{11}<0$$
$$P$$ lies outside the circle $$S=0$$ if $${S}_{11}>0$$
$$P$$ lies on the circle $$S=0,$$ if $${S}_{11}=0$$
$$\left(ii\right)$$Tangent at $$P\left({x}_{1},{y}_{1}\right)$$ on the circle $$S=0,$$ is $${S}_{1}=0$$
$$\left(iii\right)$$ Length of the tangent from the point $$\left({x}_{1},{y}_{1}\right)$$ to the circle $$S=0$$ is $$\sqrt{{S}_{11}}$$
$$\left(iv\right)$$Pair of tangents $$PQ,PR$$ from $$P\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}^{2}={S}_{11}S$$
$$\left(v\right)$$Chord of contact $$QR$$ of tangents from $$P\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}=0$$
$$\left(vi\right)$$Chord of $$S=0$$ with midpoint $$\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}={S}_{11}$$
Based on the above information,answer the following questions:

...view full instructions

The equation of the circle whose radius is $$5$$units and which touches the circle, $${x}^{2}+{y}^{2}-2x-4y-20=0$$ at the point $$\left(5,5\right)$$ is

A
$${x}^{2}+{y}^{2}+18x+16y+120=0$$

B
$${x}^{2}+{y}^{2}-18x-16y+120=0$$

C
$${x}^{2}+{y}^{2}-18x-16y-120=0$$

D
$${x}^{2}-{y}^{2}-18x-16y-120=0$$

Solution

The given circle is $${x}^{2}+{y}^{2}-2x-4y-20=0$$ has its centre
$$-2g=-2$$ or $$g=1$$
$$-2f=-4$$ or $$f=2$$$
Thus, the centre is at $$C\left(1,2\right)$$
and radius$$=r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{1}^{2}+{2}^{2}-\left(-20\right)}=5$$units
We have $${O}_{1}=\left(1,2\right)$$ and Let $${O}_{2}=\left(\alpha,\beta\right)$$ be the centres $$A\left(5,5\right)$$ is the mid-point of $${O}_{1}{O}_{2}$$
$$\therefore \dfrac{1+\alpha}{2}=5, \dfrac{2+\beta}{2}=5$$
On simplifying ,we get
$$\alpha=9,\beta=8$$
The circle is $${\left(x-9\right)}^{2}+{\left(y-8\right)}^{2}={5}^{2}$$
or $${x}^{2}+{y}^{2}-18x-16y+120=0$$
Question 3
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If the line $$y = x \sqrt{3} - 3$$ cuts the parabola $$y^2 = x + 2$$ at $$P$$ and $$Q$$ and if $$A$$ be the points $$(\sqrt{3}, 0)$$ , then $$AP.AQ$$ is

A
$$\dfrac{2}{3} (\sqrt{3} + 2)$$

B
$$\dfrac{4}{3} (\sqrt{3} + 2)$$

C
$$\dfrac{4}{3} (2 -\sqrt{3})$$

D
$$\dfrac{4}{6 - 3\sqrt{3}}$$

Solution

Parabola: $$y^2=x+2 ..........(1)$$
Line: $$y=x\sqrt{3}-3 ...........(2)$$
$$A(\sqrt3 ,0)$$ lies on $$(2)$$ and at $$x-axis$$,
$$P$$ with respect to $$A,$$ is $$P=(\sqrt3+rcos60°, rsin60°)$$
$$=(\sqrt{3} +\cfrac{r} {2},\cfrac { \sqrt { 3r } }{ 2 } )$$ (where $$r=PA)$$
$$P$$ is on Parabola,
$$=>({ \cfrac { \sqrt { 3 } r }{ 2 } })^{ 2 }=\sqrt { 3 } +\cfrac { r }{ 2 } +2$$
$$=>\cfrac { { 3r }^{ 2 } }{ 4 } =\sqrt { 3 } +2+\cfrac { r }{ 2 } $$
$$=> { 3 }r^{ 2 }-2r+4(-\sqrt { 3 } -2)=0$$
Is quadratic equation, having two roots say $$r_1$$ and $$r_2$$ which are lengths of $$PA$$ and $$AQ$$
$$AP.AQ=r_1.r_2$$
$$AP.AQ=\cfrac { 4 }{ 3 } (\sqrt { 3 } +2).$$
Question 4
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The equation $$\dfrac{x^2}{2-a}+\dfrac{y^2}{a-5}+1=0$$ represents an ellipse if $$a\in$$

A
$$\left(2,\dfrac{3}{2}\right)\cup\left(\dfrac{3}{2},5\right)$$

B
$$\left(2,\dfrac{3}{2}\right)$$

C
$$\left(1,\dfrac{3}{2}\right)$$

D
$$\left(\dfrac{3}{2},5\right)$$

Solution

Given $$\dfrac{x^2}{2-a}+\dfrac{y^2}{a-5}+1=0$$
$$\dfrac{x^2}{a-2}+\dfrac{y^2}{5-a}=1$$
For the above equation to be an ellipse
$$a-2\gt 0$$, $$5-a\gt0$$ and $$a-2 \neq 5-a$$
$$a\gt2$$, $$a\lt5$$ and $$a\neq \dfrac{7}{2}$$
$$ a \space \epsilon \left(2,\dfrac{7}{2}\right) \bigcup \left(\dfrac{7}{2},5\right)$$
Question 5
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The equation of the latusrecta of the ellipse $$9x^{2}+4^{2}-18x-8y-23=0$$ are

A
$$y=\pm \sqrt {5}$$

B
$$x=\pm \sqrt {5}$$

C
$$y=1 \pm \sqrt {5}$$

D
$$x=1 \pm \sqrt {5}$$

Solution

equation of ellipse is $${ 9x }^{ 2 }+{ 4y }^{ 2 }-18x-8y-23=0$$
$$\Rightarrow (9x^{ 2 }-18x+9)+(4y^{ 2 }-8y+4)-23-9-4=0$$
$$ \Rightarrow 9(x-1)^{ 2 }+4(y-1)^{ 2 }=36$$
$$ \Rightarrow \cfrac { (x-1)^{ 2 } }{ 4 } +\cfrac { (y-1)^{ 2 } }{ 9 } =1$$
So, equation of latus recta is $$(y-1)=\pm be$$
$$y=1\pm \sqrt { b^{ 2 }-a^{ 2 } } \Rightarrow y=1\pm \sqrt { 5 } $$
Question 6
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In the $$xy$$ plane, the segment with end points$$(3,8)$$ and $$(
5,2)$$ is the diameter of the circle. The point $$(k,10)$$ lies on the circle for:

A
no value of $$k$$

B
exactly one integral $$k$$

C
exactly one non integral $$k$$

D
two real values of $$k$$

Solution

The centre of the given circle is,
$$C\left( x,y \right)=\left( \dfrac{3+5}{2},\dfrac{8+2}{2} \right)=\left( 4,5 \right)$$

The length of the diameter of the circle is,

$$D=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( 2-8 \right)}^{2}}}=2\sqrt{10}\ units$$

Therefore,

Radius $$=\sqrt{10}\ units$$

Therefore, the equation of the required circle is,

$${{\left( x-4 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=10$$

Since, the point $$\left( k,10 \right)$$ lies on the circle, we have

$$ {{\left( k-4 \right)}^{2}}+{{\left( 10-5 \right)}^{2}}=10 $$

$$ {{k}^{2}}-8k+16+25=10 $$

$$ {{k}^{2}}-8k+31=0 $$

This equation has no real roots. Therefore, no such value of $$k$$ exists.
Question 7
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Find the equation of the circle which passes through the points $$(2,-2)$$ and $$(3,4)$$. And whose centre lies on the line $$x+y=2$$.

A
$${ x }^{ 2 }+{ y }^{ 2 }+7x+3y+16=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+4x+5y-16=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }+5x+8y+30=0$$

Solution

As $$ (2,-2)$$ passes through the circle.
$$ { x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\\ 4+4+4g-4f+c=0\\ 4g+4f+8+c=0\longrightarrow (1)$$
As $$ (3,4)$$ passes through the circle
$$ 25+6g+8f+c=0\longrightarrow (2)$$
As center of a circle $$ (-g,-f)$$
$$ -g-f=2\\ g+f=-2\longrightarrow (3)\\ 4g-4f+8+c=0\\ 4\left( g-f+2+\cfrac { c }{ 4 } \right) =0\\ g-f+2+\cfrac { c }{ 4 } =0\\ -4=\cfrac { c }{ 4 } \\ c=-16\\ 4g-4f-8=0\times 2\\ 8g-8f-16=0\\ 6g+8f+9=0\times 1\\ 6g+8f+9=0\\ 2g=-7\\ g=\cfrac { -7 }{ 2 } \\ f=\cfrac { 3 }{ 2 } $$
Equation of circle
$$ { x }^{ 2 }+{ y }^{ 2 }+2\times \cfrac { 7 }{ 2 } x-2\times \cfrac { 3 }{ 2 } y-16=0$$
$${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$$
Question 8
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The equation of the latus rectum of the parabola $${ x }^{ 2 }+4x+2y=0$$ is

A
$$2y+3=0$$

B
$$3y=2$$

C
$$2y=3$$

D
$$3y+2=0$$

Solution

Equation of Latus rectum of$${ x }^{ 2 }+4x+2y=0$$
$$\Rightarrow { x }^{ 2 }+4x+4+2y-4=0$$
$$\Rightarrow { (x+2) }^{ 2 }=-2\left( y-2 \right) $$
$$\Rightarrow -\cfrac { 1 }{ 2 } { (x+2) }^{ 2 }=(y-2)$$
$${ (x+2) }^{ 2 }=-2(y-2)$$
$$=-4\times \cfrac { 1 }{ 2 } \left( y-2 \right) $$
Latus rectum will be at a distance of $$a=\cfrac { 1 }{ 2 } $$
Distance from x-axis

$$=2-\cfrac { 1 }{ 2 } =1.5=\cfrac { 3 }{ 2 } $$
So, $$y=\cfrac { 3 }{ 2 } $$
Question 9
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A circle of radius $$2$$ lies in the first quadrant and touches both the axes of co-ordinates. Then the equation of the circle with centre $$(6, 5)$$ and touching the above circle externally is

A
$$(x - 6)^2 + (y - 5)^2 = 4$$

B
$$(x - 6)^2 + (y - 5)^2 = 9$$

C
$$(x - 6)^2 + (y - 5)^2 = 36$$

D
none of these

Solution

If $$(h,k)$$ is the center and the radius is $$r$$ then the equation of the circle is given by $$(x-h)^2+(y-k)^2=r^2$$

Given that The center of the circle $$(h,k)=(6,5)$$ and the radius $$r=2$$

Therefore, the equation of the circle is $$(x-6)^2+(y-5)^2=4$$
Question 10
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The equation $$2x^2+3y^2-8x-18y+35=\lambda$$ represents?

A
A circle for all $$\lambda$$

B
An ellipse if $$\lambda < 0$$

C
The empty set if $$\lambda > 0$$

D
A-point if $$\lambda = 0$$

Solution

Given:

$$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda $$

$$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda $$

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

For $$ \lambda = 0 $$, then

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

Thus, the point is $$ \left ( 2,3 \right ) $$.

Hence, the correct option is ‘d’.

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Conic Sections Test - 31

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Conic Sections Test - 31

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Question 1

$$(x+1)^2=16(y-2)$$

$$(x-1)^2=16(y+2)$$

$$(y-2)^2=16(x+1)$$

$$(y+2)^2=16(x-1)$$

Solution

Question 2

$${x}^{2}+{y}^{2}+18x+16y+120=0$$

$${x}^{2}+{y}^{2}-18x-16y+120=0$$

$${x}^{2}+{y}^{2}-18x-16y-120=0$$

$${x}^{2}-{y}^{2}-18x-16y-120=0$$

Solution

Question 3

$$\dfrac{2}{3} (\sqrt{3} + 2)$$

$$\dfrac{4}{3} (\sqrt{3} + 2)$$

$$\dfrac{4}{3} (2 -\sqrt{3})$$

$$\dfrac{4}{6 - 3\sqrt{3}}$$

Solution

Question 4

$$\left(2,\dfrac{3}{2}\right)\cup\left(\dfrac{3}{2},5\right)$$

$$\left(2,\dfrac{3}{2}\right)$$

$$\left(1,\dfrac{3}{2}\right)$$

$$\left(\dfrac{3}{2},5\right)$$

Solution

Question 5

$$y=\pm \sqrt {5}$$

$$x=\pm \sqrt {5}$$

$$y=1 \pm \sqrt {5}$$

$$x=1 \pm \sqrt {5}$$

Solution

Question 6

no value of $$k$$

exactly one integral $$k$$

exactly one non integral $$k$$

two real values of $$k$$

Solution

Question 7

$${ x }^{ 2 }+{ y }^{ 2 }+7x+3y+16=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x+5y-16=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+5x+8y+30=0$$

Solution

Question 8

$$2y+3=0$$

$$3y=2$$

$$2y=3$$

$$3y+2=0$$

Solution

Question 9

$$(x - 6)^2 + (y - 5)^2 = 4$$

$$(x - 6)^2 + (y - 5)^2 = 9$$

$$(x - 6)^2 + (y - 5)^2 = 36$$

none of these

Solution

Question 10

A circle for all $$\lambda$$

An ellipse if $$\lambda < 0$$

The empty set if $$\lambda > 0$$

A-point if $$\lambda = 0$$

Solution

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