JEE Main 2025 Apr 2 Shift 1, Mathematics Q16: Introduction to Ellipse

Here $a^2=18$, $b^2=9$, so $a=3\sqrt{2}$, $b=3$. Eccentricity:

$$e^2=1-\frac{b^2}{a^2}=1-\frac{1}{2}=\frac{1}{2}\Rightarrow e=\frac{1}{\sqrt{2}}$$

By the focal distance property, for $P(a\cos \theta ,b\sin \theta )$ on the ellipse:

$$SP=a-ae\cos \theta =a(1-e\cos \theta ),S^{\prime }P=a(1+e\cos \theta )$$

Hence:

$$SP\cdot S^{\prime }P=a^2(1-e^2{\cos }^2\theta )=18\left(1-\frac{{\cos }^2\theta }{2}\right)=18-9{\cos }^2\theta$$

As ${\cos }^2\theta$ runs over $[0,1]$:

$\max (SP\cdot S^{\prime }P)=18$ (at $\cos \theta =0$, i.e. ends of minor axis)

$\min (SP\cdot S^{\prime }P)=18-9=9$ (at $\cos \theta =\pm 1$, i.e. ends of major axis)

$$\min +\max =9+18=27$$

$\therefore$ Option (4) is correct.