Question

The equations of the tangents to the ellipse $4x^2+3y^2=5$ which are perpendicular to the line $3x-y+7=0$ are

• A. $2x-2y\pm \sqrt{55}=0$
• B. $2x-12y\pm \frac{\sqrt{55}}{2}=0$
• C. $2x+6y\pm \sqrt{65}=0$
• D. $2x+2y\pm \sqrt{15}=0$

Answer 1

The correct option is C $2x+6y\pm \sqrt{65}=0$

Equation of ellipse: $\frac{x^2}{\frac{5}{4}}+\frac{y^2}{\frac{5}{3}}=1$

Equation of line perpendicular to tangent: $y=3x+7\Rightarrow m_1=3$

Since lines are perpendicular $m_1\times m_2=-1\Rightarrow m_2=\frac{-1}{3}$

Equation of tangent:

$y=mx+\sqrt{a^2{m}^2+b^2}$

$\therefore y=\left(\frac{-1}{3}\right)x+\sqrt{\left(\frac{5}{4}\right)\left(\frac{1}{9}\right)+\left(\frac{5}{3}\right)}$

$\Rightarrow \frac{3y+x}{3}=\frac{\sqrt{65}}{6}$

$\Rightarrow 2x+6y\pm \sqrt{65}=0$