Q95 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
$a$ and $b$ are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes. If its latus rectum is of length 4 units and the distance between its foci is $4 \sqrt{2}$, then $a^2+b^2=$
- Correct answerA24
- B18
- C16
- D12
Solution
Length of latus rectum $=\frac{2 b^2}{a}=4$
$\Rightarrow b^2=2 a$
Distance between focii $=2 a e=2 a \sqrt{1-\frac{b^2}{a^2}}=4 \sqrt{2}$ $\Rightarrow 2 a \sqrt{1-\frac{2}{a}}=4 \sqrt{2} \Rightarrow 4 a^2\left(1-\frac{2}{a}\right)=32$ $\Rightarrow a=4, b=2 \sqrt{2} ; a^2+b^2=16+8=24$
Distance between focii $=2 a e=2 a \sqrt{1-\frac{b^2}{a^2}}=4 \sqrt{2}$ $\Rightarrow 2 a \sqrt{1-\frac{2}{a}}=4 \sqrt{2} \Rightarrow 4 a^2\left(1-\frac{2}{a}\right)=32$ $\Rightarrow a=4, b=2 \sqrt{2} ; a^2+b^2=16+8=24$
Hence, the correct answer is option: 'A'.