JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Ellipse

Let an ellipse with centre

(1, 0)

and latus rectum of length

\frac{1}{2}

have its major axis along

x

-axis. If its minor axis subtends an angle

6 0^{\circ}

at the foci, then the square of the sum of the lengths of its minor and major axes is equal to

Answer: 9

Solution

Ellipse equation:

\frac{( x - 1 ) ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

Latus rectum:

\frac{2 b ^{2}}{a} = \frac{1}{2}

⟹

4 b^{2} = a

... (i)

Minor axis subtends

6 0^{\circ}

at focus, so the angle from focus to end of minor axis is

3 0^{\circ}

tan 3 0^{\circ} = \frac{b}{a e} = \frac{1}{3}

3 b^{2} = a^{2} e^{2} = a^{2} - b^{2}

4 b^{2} = a^{2}

... (ii)

From (i) and (ii):

a = a^{2}

⟹

a = 1

b^{2} = \frac{1}{4}

⟹

b = \frac{1}{2}

(2 a + 2 b)^{2} = (2 + 1)^{2} = 9

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer