| 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. 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?

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

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.

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?

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

Answer: (B)

Solution

For

a = 2

and point of contact

(- 1, 1)

, we check which conic passes through this point. For (I)

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

1 + 1 = 2 = a^{2}

✓. The tangent to a circle at

(- 1, 1)

with equation (ii)

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

has point of contact (Q)

(\frac{- ma}{m ^{2} + 1}, \frac{a}{m ^{2} + 1})

. Setting this equal to

(- 1, 1)

\frac{- m 2}{m ^{2} + 1} = - 1

and

\frac{2}{m ^{2} + 1} = 1

. From the second equation:

2 = m^{2} + 1

, so

m^{2} = 1

and

m = \pm 1

. The first equation gives

m = 1

. Thus the combination is (I) (ii) (Q).

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

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

Solution

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