JEE 2025 — Mathematics PYQ

$\frac{x^2}{18}+\frac{y^2}{9}=1$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

On comparison, we get

a^2=18,b^2=9\Rightarrow a>b

$$PS+P{S}^{\prime }=2\times 3\sqrt{2}$$
$$b^2=a^2(1-e^2)=18(1-e^2)$$
$$\Rightarrow e=\frac{1}{\sqrt{2}}$$

% Directrix
$$\text{Directrix: }x=\frac{3\sqrt{2}}{\frac{1}{\sqrt{2}}}=6$$

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

% Max/Min Values
$$(SP\cdot S^{\prime }P{)}_{\text{max}}=18\text{ if }\cos \theta =0$$
$$(SP\cdot S^{\prime }P{)}_{\text{min}}=9\text{ if }\cos \theta =1$$
$$\text{Sum}=18+9=27$$