Let an ellipse with centre (1, 0) and latus rectum of length have its major axis along the -axis. If its minor axis subtends an angle at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________.

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Solution

Q.
Let an ellipse with centre (1, 0) and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$ -axis. If its minor axis subtends an angle $60 °$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ________. [2023]

Ans.
(9)

Length of latus rectum $= \frac{1}{2}$

and eccentricity $(a e) = \sin θ$ , where $θ$ is the angle subtended by minor axis at focus.

$\Rightarrow$ $a e = \sin 60 ° = \frac{\sqrt{3}}{2}$

Now, $\frac{2 b^{2}}{a} = \frac{1}{2} \Rightarrow b^{2} = \frac{a}{4} and a e = \frac{\sqrt{3}}{2}$

Now, $b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = a^{2} - {(a e)}^{2} = a^{2} - \frac{3}{4}$

$\Rightarrow$ $\frac{a}{4} = a^{2} - \frac{3}{4} \Rightarrow 4 a^{2} - a - 3 = 0 \Rightarrow (4 a + 3) (a - 1) = 0$

$\Rightarrow a = 1 \Rightarrow b = \frac{1}{2}$

$∴$ ${(2 a + 2 b)}^{2} = {(2 + 2 \times \frac{1}{2})}^{2} = {(3)}^{2} = 9$

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