Mathematics

Mathematics

Problem

Match 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 1Column 2Column 3
(I) x2+y2=a2x^2 + y^2 = a^2(i) my=m2x+amy = m^2x + a(P) (am2,2am)\left( \frac{a}{m^2}, \frac{2a}{m} \right)
(II) x2+a2y2=a2x^2 + a^2y^2 = a^2(ii) y=mx+am2+1y = mx + a\sqrt{m^2 + 1}(Q) (−mam2+1,am2+1)\left( \frac{-ma}{\sqrt{m^2 + 1}}, \frac{a}{\sqrt{m^2 + 1}} \right)
(III) y2=4axy^2 = 4ax(iii) y=mx+a2m2−1y = mx + \sqrt{a^2m^2 - 1}(R) (−a2ma2m2+1,1a2m2+1)\left( \frac{-a^2m}{\sqrt{a^2m^2 + 1}}, \frac{1}{\sqrt{a^2m^2 + 1}} \right)
(IV) x2−a2y2=a2x^2 - a^2y^2 = a^2(iv) y=mx+a2m2+1y = mx + \sqrt{a^2m^2 + 1}(S) (−a2ma2m2−1,−1a2m2−1)\left( \frac{-a^2m}{\sqrt{a^2m^2 - 1}}, \frac{-1}{\sqrt{a^2m^2 - 1}} \right)
For a=2a = \sqrt{2}, a tangent is drawn to a suitable conic (Column 1) at the point of contact (−1,1)(-1, 1). Which option is the only correct combination for obtaining its equation? (III) (i) (P) (II) (ii) (Q)

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