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 $\min \left(\mathrm{SP} . \mathrm{S}^{\prime} \mathrm{P}\right)+ \max \left(\mathrm{SP}^{\prime} \mathrm{S}^{\prime} \mathrm{P}\right)$ is equal to :

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Ellipse

S

and

S^{'}

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 (SP . S^{'} P) + max (SP^{'} S^{'} P)

is equal to :

Answer: (D)

For the ellipse

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

, we have

a^{2} = 18

and

b^{2} = 9

, so

a = 32

and

b = 3

.

The eccentricity:

b^{2} = a^{2} (1 - e^{2})

gives

9 = 18 (1 - e^{2})

, so

e = \frac{1}{2}

.

For any point P on the ellipse,

P S + P S^{'} = 2 a = 62

.

Using the focus-directrix property, if P has parameter

θ

P S \cdot P S^{'} = \frac{1}{2} ∣18 cos^{2} θ - 36∣

This is minimized when

cos θ = 0

(at the minor axis endpoints), giving

(P S \cdot P S^{'})_{m i n} = 9

.

It is maximized when

cos θ = \pm 1

(at the major axis endpoints), giving

(P S \cdot P S^{'})_{m a x} = 18

.

Therefore:

min (S P \cdot S^{'} P) + max (S P \cdot S^{'} P) = 9 + 18 = 27

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option