JEE Advanced 2017 · Paper 1 · Mathematics Q15

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MathematicsSingle correct+3 / -1Question 51 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) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$

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

(IV) (iii) (S)

(IV) (iv) (S)

(II) (iii) (R)

(II) (iv) (R)

Answer and worked solution

Official answer: D

Tangent. $\sqrt3x+2y=4$ means $y=-\dfrac{\sqrt3}2x+2$ , so $m=-\dfrac{\sqrt3}2$ .

Which conic? The point $(\sqrt3,\tfrac12)$ on $x^2+a^2y^2=a^2$ (II) gives $3+\dfrac{a^2}4=a^2$ , so $a=2$ : the ellipse $\dfrac{x^2}4+y^2=1$ .

Tangent form (iv). $\sqrt{a^2m^2+1}=\sqrt{3+1}=2$ . ✓

Contact point (R).
$\left(\frac{-a^2m}{\sqrt{a^2m^2+1}},\frac1{\sqrt{a^2m^2+1}}\right)=\left(\frac{2\sqrt3}2,\frac12\right)=\left(\sqrt3,\tfrac12\right).$
✓

(D)

Topics:Ellipse Tangents and normals to conics

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JEE Advanced 2017 · Paper 1, Mathematics question 15