Question

The angle between two tangents to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at the points where the line $y=1$ cuts the curve is

• A. $\frac{\pi }{4}$
• B. ${\tan }^{-1}\frac{6\sqrt{2}}{7}$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

The correct option is B ${\tan }^{-1}\frac{6\sqrt{2}}{7}$

Substituting $y=1$ in $\frac{x^2}{16}+\frac{y^2}{9}=1$
We get $x=\pm \frac{8\sqrt{2}}{3}$
And slope of tangents $\frac{dy}{dx}=\frac{9x}{16y}=m_1,m_2=\pm \frac{3\sqrt{2}}{2}$
Therefore $tan\left(\theta \right)=\frac{m_1-m_2}{1+m_1{m}_2}=|-\frac{6\sqrt{2}}{7}|\Rightarrow \theta ={tan}^{-1}\left(\frac{6\sqrt{2}}{7}\right)$