Q96 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

If the extremities of the latus recta having positive ordinate of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a \gt b)$ lie on the parabola $x^2+2 a y-4=0$, then the points $(a, b)$ lie on the curve

  • A
    $x y=4$
  • Correct answer
    B
    $x^2+y^2=4$
  • C
    $\frac{x^2}{4}+\frac{y^2}{1}=1$
  • D
    $\frac{x^2}{4}-\frac{y^2}{1}=1$

Solution

Extermities $\left( \pm a e, \frac{b^2}{a}\right)$ put in $x^2+2 a y-4=0$ $\begin{aligned} & a^2 e^2+2 b^2-4=0 ; a^2 e^2+2 a^2\left(1-e^2\right)-4=0 \\ & a^2 e^2+2 a^2-2 a^2 e^2-4=0 ; 2 a^2-a^2 e^2-4=0 \\ & \Rightarrow a^2=\frac{4}{2-e^2} ; b^2=\frac{4\left(1-e^2\right)}{\left(2-e^2\right)} \\ & a^2+b^2=4\left[\frac{1}{2-e^2}+\frac{1-e^2}{2-e^2}\right]\end{aligned}$ $a^2+b^2=4 \Rightarrow(a, b)$ satisfies $x^2+y^2=4$

Hence, the correct answer is option: 'B'.

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