Question

The equation of tangents to the ellipse $9x^2+16y^2=144$ at the ends of the latus rectum are

• A. $\sqrt{7}x+4y=16$
• B. $\sqrt{7}x-4y=16$
• C. $\sqrt{7}x-4y+16=0$
• D. $\sqrt{7}x+4y+16=0$

Answer 1

Answer: (A), (B), (C), (D)

$\sqrt{7}x+4y+16=0$

The given ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1$
Eccentricity of the ellipse is
$e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
Coordinates of the end points of the latusrectum
$(\pm ae,\pm \frac{b^2}{a})\equiv (\pm \sqrt{7},\pm \frac{9}{4})$
Hence equation of tangents are -
$\frac{x(\pm \sqrt{7})}{16}+\frac{y(\pm 9/4)}{9}=1\Rightarrow \pm \sqrt{7}x\pm 4y=16$