JEE Advanced 2017 · Paper 1 · Mathematics Q14

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MathematicsSingle correct+3 / -1Question 50 of 54

Passage

Answer Q.49, Q.50 and Q.51 by appropriately matching the information given in the three columns of the following table.

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) $my = m^2x + a$	(P) $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$
(II) $x^2 + a^2y^2 = a^2$	(ii) $y = mx + a\sqrt{m^2 + 1}$	(Q) $\left(\frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right)$
(III) $y^2 = 4ax$	(iii) $y = mx + \sqrt{a^2m^2 - 1}$	(R) $\left(\frac{-a^2m}{\sqrt{a^2m^2+1}}, \frac{1}{\sqrt{a^2m^2+1}}\right)$
(IV) $x^2 - a^2y^2 = a^2$	(iv) $y = mx + \sqrt{a^2m^2 + 1}$	(S) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$

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?

(I) (ii) (Q)

(II) (iv) (R)

(III) (i) (P)

(III) (ii) (Q)

Answer and worked solution

Official answer: C

Which conic? $(8,16)$ on $y^2=4ax$ gives $256=32a$ , so $a=8$ : the parabola (III).

Tangent. The form (i), $my=m^2x+a$ , with $m=1$ gives $y=x+8$ . ✓

Contact point. (P) gives $\left(\dfrac a{m^2},\dfrac{2a}m\right)=(8,16)$ . ✓

(C)

Topics:Parabola Tangents and normals to conics

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JEE Advanced 2017 · Paper 1, Mathematics question 14