Conic Sections Test

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Conic Sections Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The hyperbola $$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$ passes through the point $$\left( \sqrt { 6 } ,3 \right) $$ and the length of the latusrectum is $$\cfrac { 18 }{ 5 } $$. Then, the length of the transverse axis is equal to

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

E
$$1$$

Solution

Given, equation of hyperbola
$$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1...(i)\quad $$
Eq. (i) passes through $$\left( \sqrt { 6 } ,3 \right) $$
$$\therefore \cfrac { { (\sqrt { 6 } ) }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { 3 }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow \cfrac { 6 }{ { a }^{ 2 } } -\cfrac { 9 }{ { b }^{ 2 } } =1...(ii)$$
$$\cfrac { 2{ b }^{ 2 } }{ a } =\cfrac { 18 }{ 5 } ....(iii)$$
On solving Eqs. (i) and (ii)
$${ a }^{ 2 }=1,{ b }^{ 2 }=\cfrac { 9 }{ 5 } $$
Question 2
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The radius of the circle passing through the points $$(2,3),(2,7)$$ and $$(5,3)$$ is

A
$$5$$

B
$$4$$

C
$$\cfrac { 5 }{ 2 } $$

D
$$2$$

E
$$\sqrt { 5 } $$

Solution

Let the given points $$A(2,3),B(5,3)$$ and $$C(2,7)$$
Diameters of circle is distance between $$B(5,3)$$ and $$C(2,7)$$
$$d=\sqrt { { (2-5) }^{ 2 }+{ (7-3) }^{ 2 } } =\sqrt { 9+16 } $$
$$d=\sqrt { 25 } =5$$
$$\therefore$$ Radius $$=\cfrac { d }{ 2 } =\cfrac { 5 }{ 2 } $$
Question 3
1 / -0

The equation of the circumcircle of the triangle formed by the lines $$y + \sqrt{3} x = 6, y - \sqrt{3} x = 6$$ and $$y=0$$ is

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

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

C
$$x^2+y^2-4y=12$$

D
$$x^2+y^2+4x=12$$

Solution

Triangle $$ABC$$ is an equilateral triangle
Hence,
$$BC=4\sqrt{3} \\ 2R=\cfrac{a}{\sin A} \\ 2R=\cfrac{4\sqrt{3}}{\sin 60^o}=8 \\ R=4$$
Circumcentre of $$ABC$$ $$\left( \cfrac{x_1+x_2+x_3}{3},\cfrac{y_1+y_2+y_3}{3} \right) \\ =(0,2) $$
Equation of circumcircle
$$(x-0)^2+(y-2)^2=4^2 \\ x^2+y^2-4y=12$$
Question 4
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The equation of the circle having $$x-y-2=0$$ and $$x-y+2=0$$ as two tangents and $$x-y=0$$ as diameter is

A
$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y+1=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-1=0$$

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

D
$${ x }^{ 2 }+{ y }^{ 2 }=1$$

Solution

Since, the equations of tangents $$x-y-2=0$$ and $$x-y+2=0$$ are parallel.
Therefore, distance between them $$=$$ Diameter of the circle
$$=\dfrac { \left| 2-\left( -2 \right) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } }$$ $$\left( \because \dfrac { { C }_{ 2 }-{ C }_{ 1 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right) $$
$$=\dfrac { 4 }{ \sqrt { 2 } } =2\sqrt { 2 } $$
Hence, radius $$=\dfrac { 1 }{ 2 } \left( 2\sqrt { 2 } \right) =\sqrt { 2 } $$
It is clear from the figure that centre lies on the origin.
Therefore, equation of circle is $${ \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left( \sqrt { 2 } \right) }^{ 2 }$$.
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=2$$
Question 5
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The lines $$y=x+\sqrt { 2 } $$ and $$y=x-2\sqrt { 2 } $$ are the tangent of certain circle. If the point $$\left( 0,\sqrt { 2 } \right) $$ lies on this circle, then its equation is

A
$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

B
$${ \left( x+\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

C
$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y-\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

D
None of the above

Solution

Both tangents are parallel to each other, Therefore the distance between these two lines is the diameter of the circle.
diameter $$=\dfrac{3\sqrt{2}}{\sqrt{2}}=3$$
Radius is =$$\dfrac{3}{2}$$
If $$(x_1, y_1)$$ is the centre of the circle, Then

$$\dfrac{x_1-0}{1}=\dfrac{y_1-\sqrt{2}}{-1}=\dfrac{1}{2}\dfrac{|0-\sqrt{2}-2\sqrt{2}|}{2}$$

Centre is $$\bigl(\dfrac{3}{2\sqrt{2}}, \dfrac{1}{2\sqrt{2}}\Bigr)$$
Equation of the circle is $$(x-\frac{1}{3\sqrt{2}})^2+(y-\frac{1}{2\sqrt{2}})^2=9/4$$
Question 6
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On the parabola $$y={ x }^{ 2 }$$, the point least distant from the straight line $$y=2x-4$$ is

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

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

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

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

Solution

Given, parabola is $$y={ x }^{ 2 }$$ ....(i)
and straight line is $$y=2x-4$$ ....(ii)
From equations (i) and (ii), we get
$${ x }^{ 2 }-2x-4=0$$
Let $$f\left( x \right) ={ x }^{ 2 }-2x-4$$
Thus $$ f^{ ' }\left( x \right) =2x-2$$
For least distance, put $$f^{ ' }\left( x \right) =0$$
$$\Rightarrow 2x-2=0$$
$$\Rightarrow x=1$$
From equation (i), we have $$y=1$$
Hence, the point least distant from the line is $$\left( 1,1 \right) $$.
Question 7
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The curve represented by $$x = 3\cos t + 3\sin t$$ and $$y = 4\cos t - 4\sin t$$ is

A
An ellipse

B
A hyperbola

C
A parabola

D
None of these

Solution

The equations of curve are

$$x =3(\cos t + \sin t)$$

$$\Rightarrow \dfrac {x}{3} = \cos t + \sin t .... (i)$$

and $$y = 4(\cos t - \sin t)$$

$$\Rightarrow \dfrac {y}{4}= (\cos t - \sin t) .... (ii)$$

On squaring and adding (i) and (ii)

$$\dfrac {x^{2}}{9} + \dfrac {y^{2}}{16} = 2 (\sin^{2}t + \cos^{2}t)$$

$$\Rightarrow \dfrac {x^{2}}{9} + \dfrac {y^{2}}{16} = 2$$ Which is an ellipse
Hence choice (a) is correct answer.
Question 8
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The lines $$2x-3y=5$$ and $$3x-4y=7$$ are the diameters of a circle of area $$154$$ sq.units. The equation of the circle is

A
$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=62$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=47\quad $$

C
$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=47$$

D
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=62\quad $$

Solution

Given equation of lines is $$2x-3y=5$$ ..... (i) and $$3x-4y=7$$ .....(ii)
Solving above equations, we get the point of intersection as $$(1,-1)$$.
Eqns (i) and (ii) are diameters of the circle.
We know that the centre of circle $$=$$ point of intersection of diameters $$=(1,-1)$$.
Now, it is given that the area of the circle $$=154$$.
$$\pi {r}^{2}=154$$
$$\Rightarrow r=7$$
Hence, the equation of required circle is
$${(x-1)}^{2}+{(y+1)}^{2}={7}^{2}$$
$$\Rightarrow$$ $${x}^{2}+{y}^{2}-2x+2y=47$$
Question 9
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The equation of the circle which touches the lines $$x = 0, y = 0$$ and $$4x + 3y = 12$$ is

A
$$x^{2} + y^{2} - 2x - 2y - 1 = 0$$

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

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

D
$$x^{2} + y^{2} - 2x - 2y + 1 = 0$$

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

Solution

Let the radius of the circle be $$r$$.

Since, circle touches the lines $$x = 0, y = 0$$, i.e.,

$$x$$-axis and $$y$$-axis, then its centre is $$C\equiv (r, r)$$

Now, $$\text{radius} (PC) =$$ Perpendicular distance from $$(C)$$ to the line

$$4x + 3y = 12$$

i.e., $$r = \dfrac {|4r + 3r - 12|}{\sqrt {16 + 9}}$$

$$\Rightarrow 7r - 12 = \pm 5r$$

So, $$r = 1$$ or $$6$$

But $$r\neq 6$$

Required equation of circle is

$$(x - 1)^{2} + (y - 1)^{2} = 1$$

$$\Rightarrow x^{2} + 1 - 2x + y^{2} + 1 - 2y = 1$$

$$\Rightarrow x^{2} + y^{2} - 2x - 2y + 1 = 0$$
Question 10
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Equations $$x = a\cos \theta$$ and $$y= b\sin \theta$$ represent a conic section whose eccentricity $$e$$ is given by

A
$$e^{2} = \dfrac {a^{2} + b^{2}}{a^{2}}$$

B
$$e^{2} = \dfrac {a^{2} + b^{2}}{b^{2}}$$

C
$$e^{2} = \dfrac {a^{2} - b^{2}}{a^{2}}$$

D
None of these

Solution

Given equation can be rewritten as
$$\cos \theta = \dfrac {x}{a}$$ and $$\sin \theta = \dfrac {y}{b}$$
$$\because \cos^{2}\theta +\sin^{2} \theta = 1$$
$$\therefore \left (\dfrac {x}{a}\right )^{2} + \left (\dfrac {y}{b}\right )^{2} = 1$$
Thus, it represents an equation of an ellipse.
$$\therefore e = \sqrt {1 - \dfrac {b^{2}}{a^{2}}}\Rightarrow e^{2} = \dfrac {a^{2} - b^{2}}{a^{2}}$$.

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Conic Sections Test - 27

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Conic Sections Test - 27

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Question 1

$$5$$

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 2

$$5$$

$$4$$

$$\cfrac { 5 }{ 2 } $$

$$2$$

$$\sqrt { 5 } $$

Solution

Question 3

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

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

$$x^2+y^2-4y=12$$

$$x^2+y^2+4x=12$$

Solution

Question 4

$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y+1=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-1=0$$

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

$${ x }^{ 2 }+{ y }^{ 2 }=1$$

Solution

Question 5

$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

$${ \left( x+\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y-\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

None of the above

Solution

Question 6

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

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

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

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

Solution

Question 7

An ellipse

A hyperbola

A parabola

None of these

Solution

Question 8

$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=62$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=47\quad $$

$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y=47$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y=62\quad $$

Solution

Question 9

$$x^{2} + y^{2} - 2x - 2y - 1 = 0$$

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

$$x^{2} + y^{2} - 2x - 2y + 2 = 0$$

$$x^{2} + y^{2} - 2x - 2y + 1 = 0$$

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

Solution

Question 10

$$e^{2} = \dfrac {a^{2} + b^{2}}{a^{2}}$$

$$e^{2} = \dfrac {a^{2} + b^{2}}{b^{2}}$$

$$e^{2} = \dfrac {a^{2} - b^{2}}{a^{2}}$$

None of these

Solution

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