Conic Sections Test

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Conic Sections Test - 55

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Question 1
1 / -0

Consider
the set of hyperbola $$xy = {\text{ }}K,{\text{ K}} \in {\text{R,}}$$ let $${e_1}$$ be eccentricity
when $$K = \sqrt {2017} $$ and $${e_2}$$ be the
eccentricity when $$K = \sqrt {2018} $$ , then $${e_1} - {e_2}$$ is equal to

A
-1

B
0

C
2

D
1

Solution
Question 2
1 / -0

The centre of a circle is $$(2, -3)$$ and the circumference is $$10\pi$$. Then, the equation of the circle is

A
$${x}^{2}+{y}^{2}+4x+6y+12=0$$

B
$${x}^{2}+{y}^{2}-4x+6y+12=0$$

C
$${x}^{2}+{y}^{2}-4x+6y-12=0$$

D
$${x}^{2}+{y}^{2}-4x-6y-12=0$$

Solution

The circumference of the circle is given as $$10\pi $$.

$$C = 2\pi r$$

$$10\pi = 2\pi r$$

$$r = \frac{{10\pi }}{{2\pi }}$$

$$r = 5$$

The general form of the equation is,

$${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$$

The center of the circle is given as $$\left( {h,k} \right) = \left( {2, - 3} \right)$$

$${\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {\left( 5 \right)^2}$$

$${\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 25$$

$${x^2} + 4 - 4x + {y^2} + 9 + 6y = 25$$

$${x^2} + {y^2} - 4x + 6y + 13 = 25$$

$${x^2} + {y^2} - 4x + 6y - 25 + 13 = 0$$

$${x^2} + {y^2} - 4x + 6y - 12 = 0$$

Therefore, the equation of the circle is,

$${x^2} + {y^2} - 4x + 6y - 12 = 0$$
Question 3
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A conic $$C$$ passes through the points $$(2,4)$$ and is such that the segment of any of its tangents at any point contained between the co-ordinate axis is biscected at the point of tangency. Let $$S$$ denotes circle described on the foci $${F_1}$$ and $${F_2}$$ of the conic $$C$$ as diameter.
Equation of the circle $$S$$ is

A
$${x^2} + {y^2} = 16$$

B
$${x^2} + {y^2} = 8$$

C
$${x^2} + {y^2} = 32$$

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

Solution
Question 4
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The centre of a circle passing through the point $$(0,0),(1,0)$$ and touching the circle $$x^{2}+y^{2}=9$$ is ?

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

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

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

D
None of these

Solution

Let the circle be$$x^{2}+y^{2}+2gx+2fy+c=0$$
$$\because$$ It passes through $$(0,0)$$
$$\Rightarrow 0+c=0$$
$$\Rightarrow c=0$$
It also passes through $$(1,0)$$
$$\Rightarrow 1+0+2g+0+c=0$$
$$\Rightarrow g=\dfrac{-c-1}{2}=\dfrac{-1}{2}$$
$$\therefore$$Circle $$\rightarrow x^{2}+y^{2}-x+2fy=0$$
Centre$$\rightarrow (\dfrac{1}{2},-f)$$
Other circle : $$x^{2}+y^{2}=9$$
Centre \rightarrow (0,0)$$
Radius $$=3$$
$$\because Circles touches internally.
$$\therefore $$ Difference between radius = Distance between centres.
Radius of smaller circle $$=\sqrt{g^{2}+f^{2}-0}=\sqrt{\dfrac{1}{4}+f^{2}}$$
$$\Rightarrow 3-\sqrt{\dfrac{1}{4}+f^{2}}=\sqrt{(-g-0)^{2}+(-f-0)^{2}}$$
$$\Rightarrow 3-\sqrt{\dfrac{1 }{4}+f^{2}}=\sqrt{\dfrac{1}{4}+f^{2}}$$
$$\Rightarrow 2\sqrt{\dfrac{1}{4}+f^{2}}=3$$
$$\Rightarrow \dfrac{1}{4}+f^{2}=\dfrac{9}{4}$$
$$\Rightarrow f^{2}=\dfrac{8}{4}=2$$
$$\Rightarrow f=\pm \sqrt{2}$$
Hence centre $$\rightarrow (-g,-f)$$
$$\Rightarrow(\dfrac{1}{2},\pm\sqrt{2})$$
$$\therefore$$None of the options is correct.
Question 5
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If the centroid of an equilateral triangle $$(1,1)$$ and its one vertex is $$(-1,2)$$ , then the equation of the circumcircle is

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

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

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

D
$${(x + 2)^2} + {y^2} = 5$$

Solution
Question 6
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The focus of extremities of the latus rectum of the family of the ellipse $${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}{\text{ is }}\left( {b \in R} \right)$$

A
$${x^2} - ay = {a^3}$$

B
$${x^2} - ay - {e^2}$$

C
$${x^2} \pm ay = {a^2}$$

D
$${x^2} + ay - {b^2}$$

Solution

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Question 7
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Circles are drawn passing through the origin $$O$$ to intersect the coordinate axes at point $$P$$ and $$Q$$ such that $$m$$. $$PO+n.OQ=k$$, then the fixed point satisfy all of them, is given by

A
$$(m,n)$$

B
$$\dfrac{m^{2}}{k},\dfrac{n^{2}}{k}$$

C
$$\dfrac{mk}{m^{2}+n^{2}},\dfrac{nk}{m^{2}+n^{2}}$$

D
$$(k,k)$$

Solution

$${x}^{2}+{y}^{2}+2gx+2fy+c=0$$
$$C=0$$
$${x}^{2}+{y}^{2}+2gx+2fy=0$$
$${x}^{2}+2g\left(x\right)=0$$
$$x\left(x+2g\right)=0$$
$${x}^{2}+{y}^{2}+2gx+2fy=0\quad \quad H.O.P+nOQ=k$$
For$$\left(m, n\right)\quad {m}^{2}+{n}^{2}+2gm+2fn\quad \quad H\left(-2g\right)+n\left(-2f\right)=k$$
$${m}^{2}+{n}^{2}-k\neq0\quad \quad 2mg-2n7=-k$$
for$$\left(\dfrac{mk}{{m}^{2}+{n}^{2}}- \dfrac{nk}{{m}^{2}+{n}^{2}}\right)=$$
$$\left( \dfrac { { m }^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +k }^{ 2 } \right) }^{ 2 } } \quad \dfrac { n^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +n }^{ 2 } \right) }^{ 2 } } \right) +2g\left( \dfrac { mk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) +2f\left( \dfrac { nk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) \\ $$
$$=\dfrac { { k }^{ 2 }\left( { m }^{ 2 }+{ n }^{ 2 } \right) +k\left( { m }^{ 2 }+{ n }^{ 2 } \right) \left( -k \right) }{ { \left( { m }^{ 2 }+{ n }^{ 2 } \right) }^{ 2 } } $$
$$\Rightarrow \left(\dfrac{mk}{{m}^{2}+{n}^{2}}, \dfrac{nk}{{m}^{2}+{n}^{2}}\right)$$
Question 8
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The vertex $$A$$ of the parabola $${y}^{2}=4ax$$ is joined to any point $$P$$ on it and $$PQ$$ is drawn at right angles to $$AP$$ to meet the axis in $$Q$$. Projection of $$PQ$$ on the axis is equal to

A
twice the length of latus rectum

B
the latus length of rectum

C
half the length of latus rectum

D
one fourth of the length of latus rectum

Solution
Question 9
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If equation $$(5x-1)^{2}+(5y-2)^{2}=(\lambda^{2}-2\lambda+1)(3x+4y-1)^{2}$$ represents an ellipse, then $$\lambda \in$$

A
$$(0, 1)$$

B
$$(0, 2)$$

C
$$(1, 2)$$

D
$$(0, 1)\cup (1, 2)$$

Solution
Question 10
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The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes

A
$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=4$$

B
$$\dfrac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$$

C
$$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=4$$

D
none of these

Solution