| Column 1 | Column 2 | Column 3 | | --- | --- | --- | | (I) $x^{2}+y^{2}=a^{2}$ | (i) $m y=m^{2} x+a$ | (P) $\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)$ | | (II) $x^{2}+a^{2} y^{2}=a^{2}$ | (ii) $y=m x+a \sqrt{m^{2}+1}$ | (Q) $\left(\frac{-m a}{\sqrt{m^{2}+1}}, \frac{a}{\sqrt{m^{2}+1}}\right)$ | | (III) $y^{2}=4 a x$ | (iii) $y=m x+\sqrt{a^{2} m^{2}-1}$ | (R) $\left(\frac{-a^{2} m}{\sqrt{a^{2} m^{2}+1}}, \frac{1}{\sqrt{a^{2} m^{2}+1}}\right)$ | | (IV) $x^{2}-a^{2} y^{2}=a^{2}$ | (iv) $y=m x+\sqrt{a^{2} m^{2}+1}$ | (S) $\left(\frac{-a^{2} m}{\sqrt{a^{2} m^{2}-1}}, \frac{-1}{\sqrt{a^{2} m^{2}-1}}\right)$ | Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively. 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?

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Parabola

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) $y = m x + a m^{2} + 1$	(Q) $(\frac{- ma}{m ^{2} + 1}, \frac{a}{m ^{2} + 1})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + a^{2} m^{2} - 1$	(R) $(\frac{- a ^{2} m}{a ^{2} m ^{2} + 1}, \frac{1}{a ^{2} m ^{2} + 1})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + a^{2} m^{2} + 1$	(S) $(\frac{- a ^{2} m}{a ^{2} m ^{2} - 1}, \frac{- 1}{a ^{2} m ^{2} - 1})$

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

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?

(A)(I) (ii) (Q)
(B)(II) (iv) (R)
(C)(III) (i) (P)✓ Correct
(D)(III) (ii) (Q)

Answer: (C)

Solution

The tangent is

y = x + 8

with point of contact

(8, 16)

. For a parabola

y^{2} = 4 a x

, we have

1 6^{2} = 4 a (8)

, giving

256 = 32 a

and

a = 8

. The tangent equation (i) is

m y = m^{2} x + a

. With

m = 1

y = x + 8

✓. The point of contact (P) is

(\frac{a}{m ^{2}}, \frac{2 a}{m}) = (\frac{8}{1}, \frac{16}{1}) = (8, 16)

✓. Thus the combination is (III) (i) (P).

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Parabola

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