If a tangent to a suitable conic (column 1) is found to be and its point of contact is (8, 16), then which of the following options is the only correct combination? [2017]

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. If a tangent to a suitable conic (column 1) is found to be $y = x + 8$ and its point of contact is (8, 16), then which of the following options is the only correct combination?

Smart Achievers · Accepted Answer

(3)

Tangent $y = x + 8$ $\Rightarrow m = 1$ Point $(8, 16)$

$∴$ Both the coordinates as well as $m$ , are positive. The only possibility of point is $(\frac{a}{m^{2}}, \frac{2 a}{m}) = (8, 16)$ $(\frac{a}{m^{2}}, \frac{2 a}{m})$ $a = 8$

Also it satisfies the equation of curve $y^{2} = 4 a x$ for the point $(8, 16)$

And equation of tangent $m y = m^{2} x + a$ is satisfied by $m = 1$ and $a = 8$

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

Solution

1 (I)(ii)(Q)

2 (II)(iv)(R)

3 (III)(i)(P)

4 (III)(ii)(Q)

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