The tangent to a suitable conic (Column 1) at ( 3 , 1 2 ) is found to be 3 x + 2 y = 4 , then which of the following options is the only correct combination?

Question

By appropriately matching the information given in the three columns of the following table. Column 1, 2, and 3 contain conics, equations of tangents to the conics and points of contact, respectively. [2017]

	Column 1		Column 2		Column 3
(I)	$x^{2} + y^{2} = a^{2}$	(i)	$m y = m^{2} x + a$	(P)	$(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II)	$x^{2} + a^{2} y^{2} = a^{2}$	(ii)	$y = m x + a \sqrt{m^{2} + 1}$	(Q)	$(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III)	$y^{2} = 4 a x$	(iii)	$y = m x + \sqrt{a^{2} m^{2} - 1}$	(R)	$(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV)	$x^{2} - a^{2} y^{2} = a^{2}$	(iv)	$y = m x + \sqrt{a^{2} m^{2} + 1}$	(S)	$(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

Q. The tangent to a suitable conic (Column 1) at $(\sqrt{3}, \frac{1}{2})$ is found to be $\sqrt{3} x + 2 y = 4,$ then which of the following options is the only correct combination?

Smart Achievers · Accepted Answer

(4)

Point of contact $(\sqrt{3}, \frac{1}{2})$ and tangent $\sqrt{3} x + 2 y = 4$

$∴$ $m = - \frac{\sqrt{3}}{2}$

$∴$ $Both the coordinates are positive and m is negative.$

$The possibilities for points are$

$Q (\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$

or $R (\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$

For point $Q$ $(\frac{\sqrt{3} a}{\sqrt{7}}, \frac{2 a}{\sqrt{7}}) = (\sqrt{3}, \frac{1}{2})$

We get $a = \sqrt{7}$ and $a = \frac{\sqrt{7}}{4},$ which is not possible.

For point $R$ $(\frac{a^{2} \sqrt{3}}{\sqrt{3 a^{2} + 4}}, \frac{2}{\sqrt{3 a^{2} + 4}}) = (\sqrt{3}, \frac{1}{2})$

$x^{2} - a^{2} y^{2} = a^{2}$ and $\frac{2}{\sqrt{3 a^{2} + 4}} = \frac{1}{2}$

$\Rightarrow a^{4} - 3 a^{2} - 4 = 0$ and $3 a^{2} = 12$

$∴$ $a^{2} = 4$

Also for $a^{2} = 4$ , equation of ellipse

$x^{2} + a^{2} y^{2} = a^{2}$ is satisfied for the point $(\sqrt{3}, \frac{1}{2})$

$∴$ $II, (iv), R is the correct combination.$

Solution

1 (IV)(iii)(S)

2 (IV)(iv)(S)

3 (II)(iii)(R)

4 (II)(iv)(R)

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