Question

Parallel tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ such that distance between them is $\sqrt{5}$. If quadrilateral ABCD formed by these tangents, then select correct option(s)

• A. ABCD is a rhombus
• B. Acute angle between two tangents = $30^{\circ}$
• C. Area of quadrilateral ABCD is $\frac{10}{\sqrt{3}}$
• D. Area of quadrilateral ABCD is $\frac{5}{\sqrt{3}}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The standard form of the ellipse is $\frac{x^2}{2} + \frac{y^2}{1} = 1$, with $a^2=2$ and $b^2=1$. The distance between parallel tangents $y=mx \pm c$ to the ellipse is $d = \frac{2c}{\sqrt{m^2+1}}$. For the given ellipse, $c = \sqrt{a^2m^2+b^2} = \sqrt{2m^2+1}$. Thus, $d = \frac{2\sqrt{2m^2+1}}{\sqrt{m^2+1}}$. Given $d=\sqrt{5}$, we have $\sqrt{5} = \frac{2\sqrt{2m^2+1}}{\sqrt{m^2+1}}$, which leads to $m^2 = 1/3$, so $m = \pm 1/\sqrt{3}$. The constant $c = \sqrt{2(1/3)+1} = \sqrt{5/3}$. The four tangents are $y = \pm \frac{1}{\sqrt{3}}x \pm \sqrt{\frac{5}{3}}$. The vertices of the quadrilateral are $(0, \pm \sqrt{5/3})$ and $(\pm \sqrt{5}, 0)$. This forms a rhombus with diagonals $2\sqrt{5/3}$ and $2\sqrt{5}$. The area is $\frac{1}{2} \times 2\sqrt{5/3} \times 2\sqrt{5} = \frac{10}{\sqrt{3}}$. The acute angle between tangents with slopes $1/\sqrt{3}$ and $-1/\sqrt{3}$ is $60^\circ$. Therefore, options A and C are correct.