Column 1, 2 and 3 contains conics, equations of tangents to the conics and points of contact, respectively. Column 1 Column 2 Column 3 (I) x²+y²=a² (i) my=m²x+a (Q) ≤ft(a/m²,2a/m) (II) x²+a²y²=a² (ii) y=mx+a{m²+1} (Q) ≤ft({-ma}{{m²+1}},{a}{{m²+1}}) (III) y²=4ax (iii) y=mx+{a²m²-1} (R) ≤ft({-a²m}{{a²m²+1}},{1}{{a²m²+1}}) (IV) x²-a²y²=a² (iv) y=mx+{a²m²+1} (S) ≤ft({-a²m}{{a²m²-1}},{-1}{{a²m²-1}}) 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 ? [JEE Adv 2017 P1]

Question

Examsnet · Accepted Answer

Correct Answer is : (III) (i) (P)

Column 1	Column 2	Column 3
(I) $x^2+y^2=a^2$	(i) $my=m^2x+a$	$(Q) \left(\frac{a}{m^2},\frac{2a}{m}\right)$
(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) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}},\frac{-1}{\sqrt{a^2m^2-1}}\right)$

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