Question

The equation of the tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ which is parallel to the line $y=3x$, is

• A. $y=3x+\sqrt{241}$
• B. $y=3x+13$
• C. $y=3x+\sqrt{209}$
• D. none of these

Answer 1

The correct option is A $y=3x+\sqrt{241}$

Equation of line parallel to $y=3x$ is given by,
$y=3x+c..\left(1\right)$
Now this line is tangent the ellipse, $\frac{x^2}{25}+\frac{y^2}{16}=1$
$\Rightarrow c^2=a^2{m}^2+b^2=9\times 25+16=241$
$\Rightarrow c=\pm \sqrt{241}$
Thus required equation is, $y=3x\pm \sqrt{241}$