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

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

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.

The tangent to a suitable conic (Column 1) at

(3, \frac{1}{2})

is found to be

3 x + 2 y = 4

, then which of the following options is the only CORRECT combination?

(A)(IV) (iii) (S)
(B)(IV) (iv) (S)
(C)(II) (iii) (R)
(D)(II) (iv) (R)✓ Correct

Answer: (D)

Solution

The point

(3, \frac{1}{2})

and tangent

3 x + 2 y = 4

suggest an ellipse. For (II)

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

3 + a^{2} (\frac{1}{4}) = a^{2}

, giving

3 = \frac{3 a ^{2}}{4}

and

a^{2} = 4

, so

a = 2

. The tangent equation (iv) is

y = m x + a^{2} m^{2} + 1

. Rewriting the given tangent:

y = - \frac{3}{2} x + 2

, so

m = - \frac{3}{2}

. Check:

4 (\frac{3}{4}) + 1 = 4 = 2

✓. The point of contact (R) is

(\frac{- a ^{2} m}{a ^{2} m ^{2} + 1}, \frac{1}{a ^{2} m ^{2} + 1}) = (\frac{- 4 ( - \frac{3}{2} )}{2}, \frac{1}{2}) = (3, \frac{1}{2})

✓. Thus the combination is (II) (iv) (R).

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

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

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