First slide

Ellipse

Question

$Area of the quadrilateral formed by drawing tangents at the ends of latusrecta of \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1 is$

Easy

A

$\frac{16}{\sqrt{3}}$
B

$\frac{8}{\sqrt{3}}$
C

$\frac{4}{\sqrt{3}}$
D

$4 \sqrt{3}$

Solution

$\begin{array}{l} Given ellipse is \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1 \\ Here a = 2, b = 1 \end{array}$
$\begin{array}{l} e = \sqrt{\frac{a^{2} - b^{2}}{a^{2}}} = \sqrt{\frac{4 - 1}{4}} = \frac{\sqrt{3}}{2} \\ Now, end point of latusrectum = P (a e, \frac{b^{2}}{a}) \\ = P (\sqrt{3}, \frac{1}{2}) \end{array}$
$\begin{array}{l} Equation of tangent is S_{1} = 0 \\ \Rightarrow \sqrt{3} x + 2 y = 4 \end{array}$
$\begin{array}{l} Area of quadrilateral \\ = 4 \times △ A O B = 4 \times \frac{16}{2 | \sqrt{3} \times 2 |} = \frac{16}{\sqrt{3}} \end{array}$

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