JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Hyperbola

The tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$ has the form $y=mx\pm \sqrt{a^2{m}^2-16}$. Comparing with $y=2x+1$ (rewriting $2x-y+1=0$), we have $m=2$ and $\sqrt{a^2{m}^2-16}=1$. This gives $4a^2-16=1$, so $a^2=\frac{17}{4}$ and $a=\frac{\sqrt{17}}{2}$. Now $2a=\sqrt{17}$. Checking which can be sides of a right triangle: $(2a{)}^2+4^2=17+16=33\ne 1$, so (D) works: $17+16=33$ and $1^2=1$, but $33\ne 1$. Actually $(2a{)}^2=17$, $4^2=16$, $1^2=1$; we need $17=16+1$, which is true. So (D) CAN be sides. For (A): $a^2+4^2=\frac{17}{4}+16=\frac{81}{4}$ and $1^2=1$, not equal. For (B): $a^2+4^2=\frac{17}{4}+16=\frac{81}{4}$ and $2^2=4$, not equal. For (C): $(2a{)}^2+8^2=17+64=81$ and $1^2=1$, not equal. Options (A), (B), (C) cannot be sides of right triangles.