Question

Tangents are drawn to the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ at the end points of the latus rectum. The area of quadrilateral formed by these tangents is

• A. $\frac{128}{3}$ sq. units
• B. $\frac{32}{3}$ sq. units
• C. $\frac{64}{3}$ sq. units
• D. $\frac{256}{3}$ sq. units

Answer 1

The correct option is A $\frac{128}{3}$ sq. units

For ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1,$
$e=\sqrt{1-\frac{7}{16}}=\frac{3}{4}$

One end of latus rectum is
$(ae,\frac{b^2}{a})\equiv (3,\frac{7}{4})$
Therefore, equation of tangents at this point is

$\frac{3x}{16}+\frac{7}{4}.\frac{y}{7}=1$
$\Rightarrow \frac{3x}{16}+\frac{y}{4}=1$
It meets the coordinate axes at $(\frac{16}{3},0)$ and $(0,4).$
Hence, the required area of quadrilateral
$=4\times \text{area of}△AOB$
$=4\times (\frac{1}{2}\times \frac{16}{3}\times 4)=\frac{128}{3}$ sq. units