Conic Sections Test

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Conic Sections Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the circle whose centre and radius are $$\left( 1,-1 \right) $$ and $$4$$ respectively, is

A
$${ x }^{ 2 }-{ y }^{ 2 }+2x+2y-14=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-14=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+14=0$$

Solution

We know that equation of circle is
$${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$$
Here, centre is $$\left( 1,-1 \right) $$ and radius is $$4$$.
$$\therefore $$ Equation is
$${ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }={ \left( 4 \right) }^{ 2 }$$
$$\Rightarrow { x }^{ 2 }+1-2x+{ y }^{ 2 }+1+2y=16$$
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$$
Question 2
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For hyperbola $$\dfrac{x^2}{cos^2a}-\dfrac{y^2}{sin^2a}=1$$ which of the following remains constant with change in 'a'?

A
Abscissae of vertices

B
Abscissae of foci

C
Eccentricity

D
Directrix

Solution

Comparing given equation $$\dfrac{x^2}{cos^2a}-\dfrac{y^2}{sin^2a}=1$$ with the standard equation $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} =1$$, we get
$$a^2 = cos^2 a$$ and $$b^2 = sin^2 a$$
$$\therefore 1=sin^2 a + cos^2 a = a^2 + b^2$$
$$\Rightarrow e=\sqrt { \dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 } } } $$
$$\Rightarrow e=\dfrac{1}{cos\,a}$$
Now, foci $$ae=cos\,a.\dfrac{1}{cos\,a}=1$$
Question 3
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The distance between the foci of the hyperbola $${ x }^{ 2 }-3{ y }^{ 2 }-4x-6y-11=0$$ is

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

Given, equation of hyperbola is
$${ x }^{ 2 }-3{ y }^{ 2 }-4x-6y-11=0$$
$$\Rightarrow \left( { x }^{ 2 }-4x+4 \right) -3\left( { y }^{ 2 }+2y+1 \right) -11=4-3$$
$$\Rightarrow { \left( x-2 \right) }^{ 2 }-3{ \left( y+1 \right) }^{ 2 }=12$$
$$\Rightarrow \dfrac { { \left( x-2 \right) }^{ 2 } }{ 12 } -\dfrac { { \left( y+1 \right) }^{ 2 } }{ 4 } =1$$
Now, $$e=\sqrt { 1+\dfrac { 4 }{ 12 } } =\dfrac { 2 }{ \sqrt { 3 } } $$
$$\therefore $$ Distance between foci $$=2ae=2\times \sqrt { 12 } \times \dfrac { 2 }{ \sqrt { 3 } } =8$$
Question 4
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The equation of the ellipse having vertices at $$\displaystyle \left( \pm 5,0 \right) $$ and foci $$\displaystyle \left( \pm 4,0 \right) $$ is

A
$$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1$$

B
$$\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$$

C
$$\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$$

D
$$\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$$

Solution

The vertices and foci of an ellipse are $$\displaystyle \left( \pm 5,0 \right) $$ and $$\displaystyle \left( \pm 4,0 \right) $$ respectively.
$$\displaystyle \therefore \quad a=5$$ and $$\displaystyle ae=4$$
$$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 } $$
We know that,
$$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } } $$
$$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$$
Hence, equation of an ellipse is
$$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$$
Question 5
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The equation of a parabola which passes through the intersection of a straight line x$$+$$y$$=0$$ and the circle $$x^2+y^2+4y=0$$ is.

A
$$y^2=4x$$

B
$$y^2=x$$

C
$$y^2=2x$$

D
None of these

Solution

$$x+y=0$$
$$x^2+y^2+4y=0$$
The line and the circle intersect at the point $$(0,0)$$ and $$(2,-2)$$
Among the given options, only $$y^2=2x$$ passes through both these points.

So, the answer is option (C).
Question 6
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The equation to the circle with centre $$(2, 1)$$ and touching the line $$3x + 4y = 5$$ is

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

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

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

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

Solution

The distance from the line $$ax+by+c=0$$ from the line to a point $$(x_0,y_0)$$ is:
$$D=\dfrac { \left| a{ x }_{ 0 }+b{ y }_{ 0 }+c \right| }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } }$$
Here, the distance is equal to the radius of the circle and the distance from the line $$3x+4y-5=0$$ to a point $$(2,1)$$ is:
$$r=\dfrac { \left| (3\times 2)+(4\times 1)+(-5) \right| }{ \sqrt { 3^{ 2 }+{ 4 }^{ 2 } } } =\dfrac { \left| 6+4-5 \right| }{ \sqrt { 9+16 } } =\dfrac { \left| 5 \right| }{ \sqrt { 25 } } =\dfrac { 5 }{ 5 } =1$$
Therefore, equation of the circle with centre $$(2,1)$$ and radius $$1$$ is:
$${ \left( x-2 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( 1 \right) }^{ 2 }\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1-2y=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+5=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+4=0$$
Hence, the equation of circle is $$x^2+y^2-4x-2y+4=0$$.
Question 7
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Find the equation of a circle with center $$(2,0)$$ and passing through point $$\left( 3,\sqrt { 3 } \right) $$.

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

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

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

D
$$(x + 2)^{2} + y^{2} = 16$$

Solution

Let the radius be '$$r$$'.
Then the equation of the circle will be
$$(x-2)^{2}+y^{2}=r^{2}$$
Now, substituting the point $$(3,\sqrt{3})$$ in the above equation gives us $$(3-2)^{2}+(\sqrt{3})^{2}=r^{2}$$
$$\Rightarrow 1+3^{2}=r^{2}$$ or $$r^{2}=4$$
Hence, the equation of the circle is $$(x-2)^{2}+y^{2}=4$$.
Question 8
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If $$(a,b)$$ lies on circle with centre as origin, then its radius will be

A
$$\displaystyle a-b$$

B
$$\displaystyle a+b$$

C
$$\displaystyle \sqrt { { a }^{ 2 }+{ b }^{ 2 } } $$

D
$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }$$

Solution

We know the formula,

The equation of a circle of radius r and centre the origin is

$$x^2+y^2=r^2$$

Here the center is (a, b)

so Radius , $$r = \sqrt{a^2+b^2}$$
Question 9
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The radius of the circle $$x^{2} + y^{2} + 4x + 6y + 13 = 0$$ is

A
$$\sqrt {26}$$

B
$$\sqrt {13}$$

C
$$\sqrt {23}$$

D
$$0$$

Solution

Given equation is
$$x^{2} + y^{2} + 4x + 6y + 13 = 0$$
or $$(x^{2} + 4x + 4) + (y^{2} + 6y + 9) + 13 = 4 + 9$$
or $$(x + 2)^{2} (y + 3)^{2} = 0$$
$$\therefore$$ Radius of circle $$= 0$$.
Question 10
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The sum of the focal distances of any point on the conic $$\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$$ is

A
$$10$$

B
$$9$$

C
$$41$$

D
$$18$$

Solution

We know, if P is any point on the curve, then Sum of focal distances $$=$$ length of major axis
i.e., $$SP + S'P = 2a$$
$$= 2(5) [\because a^{2} = 5^{2}]$$
$$= 10$$

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Conic Sections Test - 22

Result

Conic Sections Test - 22

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SHARING IS CARING

Question 1

$${ x }^{ 2 }-{ y }^{ 2 }+2x+2y-14=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-14=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+14=0$$

Solution

Question 2

Abscissae of vertices

Abscissae of foci

Eccentricity

Directrix

Solution

Question 3

$$4$$

$$6$$

$$8$$

$$10$$

Solution

Question 4

$$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1$$

$$\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$$

$$\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$$

$$\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$$

Solution

Question 5

$$y^2=4x$$

$$y^2=x$$

$$y^2=2x$$

None of these

Solution

Question 6

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

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

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

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

Solution

Question 7

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

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

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

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

Solution

Question 8

$$\displaystyle a-b$$

$$\displaystyle a+b$$

$$\displaystyle \sqrt { { a }^{ 2 }+{ b }^{ 2 } } $$

$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }$$

Solution

Question 9

$$\sqrt {26}$$

$$\sqrt {13}$$

$$\sqrt {23}$$

$$0$$

Solution

Question 10

$$10$$

$$9$$

$$41$$

$$18$$

Solution

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