For a = 2 , if a tangent is drawn to a suitable conic (Column 1) at the point of contact ( - 1 , 1 ) , then which of the following options is the only correct combination for obtaining its equation?

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. For $a = \sqrt{2}$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact $(- 1, 1)$ , then which of the following options is the only correct combination for obtaining its equation?

Smart Achievers · Accepted Answer

(2)

For $a = \sqrt{2}$ and point of contact $(- 1, 1)$

Equation of circle is satisfied

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

then equation of tangent is

$- x + y = 2$ $\Rightarrow m = 1$ and point of contact

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

$∴$ $(I), (i i), (Q) is the correct combination.$

Solution

1 (I)(i)(P)

2 (I)(ii)(Q)

3 (II)(ii)(Q)

4 (III)(i)(P)

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