IIT-JEE 2012 · Paper 1, Mathematics question 54

Tangents of slope 2. For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $y=mx+c$ is a tangent when $c^2=a^2m^2-b^2=36-4=32$, so $c=\pm4\sqrt2$.

Point of contact of $y=mx+c$: $\left(-\dfrac{a^2m}c,-\dfrac{b^2}c\right)$.

• $c=4\sqrt2$: $\left(-\dfrac{18}{4\sqrt2},-\dfrac4{4\sqrt2}\right)=\left(-\dfrac9{2\sqrt2},-\dfrac1{\sqrt2}\right)$.
• $c=-4\sqrt2$: $\left(\dfrac9{2\sqrt2},\dfrac1{\sqrt2}\right)$.

(A), (B)