JEE Advanced 2017 · Paper 1 · Mathematics Q13

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MathematicsSingle correct+3 / -1Question 49 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) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$

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?

(I) (i) (P)

(I) (ii) (Q)

(II) (ii) (Q)

(III) (i) (P)

Answer and worked solution

Official answer: B

Which conic? With $a=\sqrt2$ , $(-1,1)$ lies on $x^2+y^2=2$ (I).

Tangent. The slope form for a circle is (ii), $y=mx+a\sqrt{m^2+1}$ , with point of contact (Q), $\left(\dfrac{-ma}{\sqrt{m^2+1}},\dfrac a{\sqrt{m^2+1}}\right)$ .

Check. $m=1$ gives contact $\left(-\tfrac{\sqrt2}{\sqrt2},\tfrac{\sqrt2}{\sqrt2}\right)=(-1,1)$ . ✓

(B)

Topics:Tangents and normals Tangents and normals to conics

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JEE Advanced 2017 · Paper 1, Mathematics question 13