Conic Sections Test

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Conic Sections Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The latus rectum of the parabola $$\displaystyle x = at^2 + bt + c, y = a't^2 + b't + c'$$ is

A
$$\displaystyle \dfrac {\left ( aa'- bb' \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$$

B
$$\displaystyle \dfrac {\left ( ab'- a'b \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$$

C
$$\displaystyle \dfrac {\left ( bb'- aa' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$$

D
$$\displaystyle \dfrac {\left ( a'b- ab' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$$

Solution
Question 2
1 / -0

The centre of a circle is $$C(2,-5)$$ and the circle passes through the point $$A(3,2)$$. The equation of the circle is

A
$${ x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-21=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+4x-10y+21=0$$

D
none of these

Solution

radius of the circle is $$AC=\sqrt { { (3-2) }^{ 2 }+{ (2+5) }^{ 2 } } =\sqrt { 1+49 } =\sqrt { 50 } $$
equation of the circle is
$${ (x-x_1) }^{ 2 }+{ (y-y_1) }^{ 2 }=r^2$$
$${ (x-2) }^{ 2 }+{ (y+5) }^{ 2 }=50\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$
Question 3
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If the parabola $${y}^{2}=4ax$$ passes through the point $$P(3,2)$$, then the length of its latus rectum is

A
$$\cfrac{1}{3}$$

B
$$\cfrac{2}{3}$$

C
$$\cfrac{4}{3}$$

D
$$4$$

Solution

Since the point $$P(3,2)$$ lies on $${ y }^{ 2 }=4ax$$, we have
$$4a\times 3={ 2 }^{ 2 }\Rightarrow a=\cfrac { 1 }{ 3 } $$
Latus rectum $$=4a=4\times \cfrac { 1 }{ 3 } =\cfrac { 4 }{ 3 } $$
Question 4
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The equation $$\displaystyle ax^2 + 4xy + y^2 + ax + 3y + 2 = 0$$ represents a parabola if a is

A
$$\displaystyle -4$$

B
$$\displaystyle 4$$

C
$$\displaystyle 0$$

D
$$\displaystyle 8$$

Solution
Question 5
1 / -0

The foci of the ellipse $$25\left ( x + 1 \right )^2 + 9\left ( y + 2 \right )^2 = 225$$, are at

A
$$\left ( -1 , 2 \right )$$ and $$\left ( -1 , -6 \right )$$

B
$$\left ( -2 , 1 \right )$$ and $$\left ( -2 , 6 \right )$$

C
$$\left ( -1 , -2 \right )$$ and $$\left ( -2 , -1 \right )$$

D
$$\left ( -1 , -2 \right )$$ and $$\left ( -1 , -6 \right )$$

Solution
Question 6
1 / -0

The latus rectum of the hyperbola $$9x^{2} - 16y^{2} - 18x - 32y - 151 = 0$$ is

A
$$\dfrac{9}{4}$$

B
9

C
$$\dfrac{3}{2}$$

D
$$\dfrac{9}{2}$$

Solution

Hyperbola $$9x^{2} - 16y^{2} - 18x - 32y - 151 = 0$$ can be written as

$$9(x^{2}-2x) - 16 (y^{2}+2y) = 151$$

$$\Rightarrow 9(x-1)^{2} - 16(y+1)^{2} = 151 + 9 - 16 = 144$$

$$\Rightarrow \dfrac{(x-1)^{2}}{16} - \dfrac{(y+1)^{2}}{9} = 1$$

or $$\dfrac{X^{2}}{16} - \dfrac{Y^{2}}{9} = 1$$

$$\left [ where X = x-1, Y= y+1 \right ]$$

Here $$a^{2} = 16, b^{2} = 9$$

Latus rectum $$= 2\dfrac{b^{2}}{a} = \dfrac{2(9)}{4} = \dfrac{9}{2}$$
Question 7
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If the equation of the ellipse is $$3x^{2}+ 2 y^{2}+6x-8y+5=0, $$ then which of the following is/ are true?

A
$$e=\dfrac{1}{\sqrt{3}}$$

B
center is (-1,2)

C
foci are (-1,1) and (-1,3)

D
directrices are $$y= 2 \pm \sqrt{3}$$

Solution

$$3x^{2}+ 2y^{2}+ 6x- 8y+ 5=0$$
$$\dfrac{(x+1)^{2}}{2}+ \dfrac{(y-2)^{2}}{3}=1 $$
Therefore, center is (-1,2) and ellipse is vertical (b>a)
$$a^{2}=2, b^{2}=3$$
Now $$2=3(1-e^{2})$$
$$e=\dfrac{1}{\sqrt{3}}$$
Foci are(-1,2$$ \pm \, be)\, and \, (-1,2 \pm 1)= (-1,3)\, and\, (-1,1)$$ and directrix are $$y=2 \pm \dfrac{b}{e} \implies y=5\, and \, y=-1$$
Question 8
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Consider the parabola whose focus is at (0,0) and tangent at vertex is $$ x-y+1=0 $$

The length of latus rectum is

A
$$ 4 \sqrt{2} $$

B
$$ 2 \sqrt{2} $$

C
$$ 8 \sqrt{2} $$

D
$$ 3 \sqrt{2} $$

Solution

The distance between the focus and the tangent at the
$$\text { vertex }=\dfrac{|0-0+1|}{\sqrt{1^{2}+1^{2}}}=\dfrac{1}{\sqrt{2}}$$
The directrix is the line parallel to the tangent at vertex and at a distance $$ 2 \times \dfrac{1}{\sqrt{2}} $$ from the focus.
Let equation of directrix is
$$x-y+\lambda=0$$
$$\text { where } \quad \dfrac{\lambda}{\sqrt{1^{2}+1^{2}}}=\dfrac{2}{\sqrt{2}} $$
$$\Rightarrow \quad \lambda=2$$
Let $$ P(x, y) $$ be any moving point on the parabola, then
$$O P =P M $$
$$\Rightarrow x^{2}+y^{2} =\left(\dfrac{x-y+2}{\sqrt{1^{2}+1^{2}}}\right)^{2}$$
$$ \Rightarrow 2 x^{2}+2 y^{2}=(x-y+2)^{2}$$
$$\Rightarrow x^{2}+y^{2}+2 x y-4 x+4 y-4=0$$
Latus rectum length $$ =2 \times $$ (distance of focus from directrix )
$$=2\left|\dfrac{0-0+2}{\sqrt{1^{2}+1^{2}}}\right| $$
$$=2 \sqrt{2}$$
Solving parabola with $$ x $$ -axis,
$$x^{2}-4 x-4 =0 $$
$$\Rightarrow x =\dfrac{4 \pm \sqrt{32}}{2}=2 \pm 2 \sqrt{2}$$
$$ \Rightarrow $$ Length of chord on $$ x $$ -axis is $$ 4 \sqrt{2} $$
Since the chord $$ 3 x+2 y=0 $$ passes through the focus, it is focal chord.
Hence, tangents at the extremities of chord are perpendicular.
Question 9
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The length of the latus rectum of the parabola whose focus is $$\left (\frac{u^{2}} {2g} \sin 2\alpha, -\frac{u^{2}} {2g} \cos 2 \alpha \right )$$ and directrix is $$y = \frac{u^{2}} {2g}$$ is

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

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

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

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

Solution

Focus of parabola is $$\left (\frac{u^{2}} {2g} \sin 2\alpha, -\frac{u^{2}} {2g} \cos 2 \alpha \right )$$ and directrix is $$y = \frac{u^{2}} {2g}$$
Length of latus rectum
$$ = 2 \times$$ distance of focus form directrix
$$ = 2 \times \left |\dfrac{-\frac{u^{2}} {2g} \cos 2 \alpha - \frac{u^{2}} {2g}} {\sqrt{1}} \right | $$
$$= \dfrac{2u^{2}} {g} \cos^{2} \alpha$$
Question 10
1 / -0

Directions For Questions

A curve is represented by C = $$21 x^{2}- 6xy+ 29y^{2}+ 6x- 58y- 151=0$$

...view full instructions

The lengths of axes

A
$$6, 2\sqrt{6}$$

B
$$5, 2 \sqrt{5}$$

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

D
none of these

Solution

$$21x^{2}- 6xy+ 29y^{2}+ 6x-58y-151=0$$
$$3(x-3y+3)^{2}+ 2(3x+y-1)^{2}=180$$
$$\dfrac{(x-3y+3)^{2}}{60}+ \dfrac{(3x+y-1)^{2}}{90}=1 $$
$$(\dfrac{x-3y+3}{\sqrt{1+3^{2}}\sqrt{6}})^{2}+ (\dfrac{3x+y-1}{\sqrt{1+ 3^{2}3}})^{2}=1 $$
Thus C is an ellipse whose lengths of axes are $$6, 2\sqrt{6}$$

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Conic Sections Test - 50

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Conic Sections Test - 50

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SHARING IS CARING

Question 1

$$\displaystyle \dfrac {\left ( aa'- bb' \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$$

$$\displaystyle \dfrac {\left ( ab'- a'b \right )^2}{\left ( a^2 + a'^2 \right )^\dfrac {3}{2}}$$

$$\displaystyle \dfrac {\left ( bb'- aa' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$$

$$\displaystyle \dfrac {\left ( a'b- ab' \right )^2}{\left ( b^2 + b'^2 \right )^\dfrac {3}{2}}$$

Solution

Question 2

$${ x }^{ 2 }+{ y }^{ 2 }-4x+10y-21=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-21=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x-10y+21=0$$

none of these

Solution

Question 3

$$\cfrac{1}{3}$$

$$\cfrac{2}{3}$$

$$\cfrac{4}{3}$$

$$4$$

Solution

Question 4

$$\displaystyle -4$$

$$\displaystyle 4$$

$$\displaystyle 0$$

$$\displaystyle 8$$

Solution

Question 5

$$\left ( -1 , 2 \right )$$ and $$\left ( -1 , -6 \right )$$

$$\left ( -2 , 1 \right )$$ and $$\left ( -2 , 6 \right )$$

$$\left ( -1 , -2 \right )$$ and $$\left ( -2 , -1 \right )$$

$$\left ( -1 , -2 \right )$$ and $$\left ( -1 , -6 \right )$$

Solution

Question 6

$$\dfrac{9}{4}$$

9

$$\dfrac{3}{2}$$

$$\dfrac{9}{2}$$

Solution

Question 7

$$e=\dfrac{1}{\sqrt{3}}$$

center is (-1,2)

foci are (-1,1) and (-1,3)

directrices are $$y= 2 \pm \sqrt{3}$$

Solution

Question 8

$$ 4 \sqrt{2} $$

$$ 2 \sqrt{2} $$

$$ 8 \sqrt{2} $$

$$ 3 \sqrt{2} $$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$6, 2\sqrt{6}$$

$$5, 2 \sqrt{5}$$

$$4, 4 \sqrt{5}$$

none of these

Solution

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