Question

The line $3x+5y=15\sqrt{2}$ is a tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ at a point whose eccentric angle is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{6}$

REF.IMAGE

$3x+5y=15\sqrt{2}$

$x=\frac{15\sqrt{2}-5y}{3}$

$\frac{x^{252}}{25}+\frac{y^2}{9}=1\Rightarrow \frac{(\frac{15\sqrt{2}-5y}{3}{)}^2}{25}+\frac{y^2}{9}=1$

$\Rightarrow (3\sqrt{2}-y{)}^2+y^2=9$

$\Rightarrow 18+y^2-6\sqrt{2}y+y^2=9$

$\Rightarrow 2y^2-6\sqrt{2}y+9=0$

$2y^2-3\sqrt{2}y-3\sqrt{2y}+9=0$

$\sqrt{2}y(\sqrt{2}y-3)-3(\sqrt{2}y-3)=0$

$y=\frac{3}{\sqrt{2}}\Rightarrow \frac{x^2}{25}+\frac{9}{2\times 9}=1$

$\frac{x^2}{25}=1/2\Rightarrow x=\frac{5}{\sqrt{2}}$

$\therefore (\frac{1}{\sqrt{2}},\frac{5}{\sqrt{2}})$

$tan\theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =30$