Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively. Column 1 Column 2 Column 3 (I) x 2 + y 2 = a 2 (i) m y = m 2 x + a (P) a m 2 , 2 a m (II) x

Question

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

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

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?

TestHub · Accepted Answer

y = x + 8 is tangent ⇒ m = 1 ; P 8 , 16 Comparing tangent with (i) of column- 2 , m = 1 satisfied and a = 8 is obtained which matches for point of contact (P) of column 3 and (III) of column- I.

Mathematics - Parabola Question with Solution | TestHub

Options:

Doubts & Discussion