Question

The equation of tangents to the ellipse $ \frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1 $ , which are perpendicular to the line $ 3x+4y=7 $ , are

• A. $ 4x-3y=\pm \sqrt{20} $
• B. $ 4x-3y=\pm \sqrt{12} $
• C. $ 4x-3y=\pm \sqrt{2} $
• D. $ 4x-3y=\pm 1 $

Answer 1

Answer: (A)

Let the equation of tangent, which is perpendicular to the line
$ 3x+4y=7 $ , is $ 4x-3y=\lambda $ .
Since, it is a tangent to the ellipse.
$ \therefore $ $ {{\lambda }^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}} $
Here, $ {{a}^{2}}=9,\,\,{{b}^{2}}=4 $ and $ m=\frac{4}{3} $
$ \therefore $ $ {{\lambda }^{2}}=9\times {{\left( \frac{4}{3} \right)}^{2}}+4 $
$ =16+4 $
$ \Rightarrow $ $ {{\lambda }^{2}}=20 $
$ \Rightarrow $ $ \lambda =\pm \sqrt{20} $
$ \therefore $ Equation is $ 4x-3y=\pm \sqrt{20} $