Question

$\mathrm{The}\mathrm{value}\left(s\right)\mathrm{of}\lambda \mathrm{for}\mathrm{which}\mathrm{the}\mathrm{line\; y=2x+\lambda }$touches the hyperbola $16x^2-9y^2=144$is/are:

• A. $\sqrt{5}$
• B. -$\sqrt{5}$
• C. $2\sqrt{5}$
• D. $-2\sqrt{5}$

Answer 1

Answer: (C), (D)

The correct options are
C $2\sqrt{5}$
D $-2\sqrt{5}$

$\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{hyperbola}\mathrm{is}$
$16x^2-9y^2=144\mathrm{or}\frac{x^2}{9}-\frac{y^2}{16}=1\mathrm{On}\mathrm{comparing}\mathrm{with}\mathrm{the}\mathrm{standard}\mathrm{form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\mathrm{we}\mathrm{get}:a^2=9,b^2=16\mathrm{and}\mathrm{On}\mathrm{comparing}y=2x+\lambda \mathrm{with}\mathrm{the}\mathrm{slope}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{tangent}y=\mathrm{mx}\pm \sqrt{a^2{m}^2-b^2},\mathrm{we}\mathrm{get}-m=2\mathrm{and}\sqrt{a^2{m}^2-b^2}=\pm \lambda \Rightarrow {\lambda }^2=9(2{)}^2-16=36-16=20\lambda =\pm 2\sqrt{5}$