Exam

JEE Main

Year

2025

Date

2 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Ellipse

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Ellipse

If S and S^ are the foci of the ellipse (x^2)/(18)+(y^2)/(9)=1 and...

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and P be a point on the ellipse, then $\min \left(S P . S^{\prime} \mathrm{P}\right)+$ $\max \left(\mathrm{SP} . \mathrm{S}^{\prime} \mathrm{P}\right)$ is equal to :

Step-by-step solution

$\mathrm{PS}+\mathrm{PS}^{\prime}=2 \times 3 \sqrt{2}$
$\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \Rightarrow 9=18\left(1-\mathrm{e}^2\right)$
$\Rightarrow \mathrm{e}=\frac{1}{\sqrt{2}}$
Directrix $\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=\frac{3 \sqrt{2}}{\frac{1}{\sqrt{2}}}=6$
$P S \cdot P S '=\left|\frac{1}{\sqrt{2}}(3 \sqrt{2} \cos \theta-6) \frac{1}{\sqrt{2}}(3 \sqrt{2} \cos \theta+6)\right|$
$=\frac{1}{2}\left|18 \cos ^2 \theta-36\right|$
\left(\mathrm{PS} \cdot \mathrm{PS}^{\prime}\right)_{\max }=18 ;(\mathrm{PS} \cdot \mathrm{PS})_{\min }=9$<br/>sum$=27