Question

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 60° at the foci, then the square of the sum of the lengths of its minor and major axes is equal to

Answer 1

Correct Answer: 9

The given ellipse has its center at (1, 0) with its major axis along the x-axis, and a latus rectum of length   \(\frac{1}{2}\). The standard form of an ellipse with the major axis along the x-axis is:
\[ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \]
Here, \((h,k)=(1,0)\), and the latus rectum \( \) of an ellipse is given by \(\frac{2b^2}{a}\), where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. Given the length of the latus rectum is \(\frac{1}{2}\):
\[ \frac{2b^2}{a}=\frac{1}{2} \] \[ 4b^2=a \quad \text{(1)} \]
The minor axis subtends an angle of 60° at the foci. The distance from the center to each focus is \(c=\sqrt{a^2-b^2}\). The angle subtended by the minor axis at the focus \(2\theta=60^\circ\) implies \( \theta = 30^\circ\). Using trigonometry, the minor axis length is:
\[ 2b = 2c\tan(30^\circ) \] \[ b = c\tan(30^\circ) = c\cdot \frac{1}{\sqrt{3}} \]
Substitute \(c=\sqrt{a^2-b^2}\):
\[ b=\frac{\sqrt{a^2-b^2}}{\sqrt{3}} \quad \text{(2)} \]
Using (1) and (2):
\[ a = 4b^2 \] \[ b = \frac{\sqrt{16b^4-b^2}}{\sqrt{3}} \]
Solve:
\[ b^2 = \frac{16b^4}{3} - \frac{b^2}{3} \] \[ 3b^2 = 16b^4 - b^2 \] \[ 3b^2 = 16b^4 - b^2 \] \[ 16b^4 - 4b^2 = 0 \] \[ 4b^2(4b^2 - 1) = 0 \]
Either \(4b^2=0\) or \(4b^2=1\), giving \(b^2=\frac{1}{4}\)
Compute \(a^2\):
\[ a^2=4b^2=\frac{1}{1} \quad \Rightarrow a=1 \quad b=\frac{1}{2} \]
The sum of the lengths of the axes is \(2a+2b=2(1)+2\left(\frac{1}{2}\right)=3\)
The square of the sum is:
\[ \left(3\right)^2=9 \]