Q96 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1,(b \lt 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is 8 and $\mathbf{S P}=7$, then $\theta=$

  • A
    $\frac{\pi}{6}$
  • Correct answer
    B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{2 \pi}{3}$

Solution

$\begin{aligned} & \text { } 2 a c=8 \\ & S^{\prime} p=e\left(\frac{a}{e}+2\right) ; 7=a+e x=7 \\ & a e=4 ; e=\frac{4}{5} ; S^{\prime} P=7=5+\frac{4}{5} x \Rightarrow x=2.5 \\ & a \cos \theta=2.5 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} .\end{aligned}$

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

Relevant Questions