Question ID: #597
If $S$ 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 $\min(SP \cdot S’P) + \max(SP \cdot S’P)$ is equal to:
- (1) $3(1+\sqrt{2})$
- (2) $3(6+\sqrt{2})$
- (3) 9
- (4) 27
Solution:
For the ellipse $\frac{x^2}{18} + \frac{y^2}{9} = 1$:
$$a^2 = 18, \quad b^2 = 9$$
$$b^2 = a^2(1-e^2) \Rightarrow 9 = 18(1-e^2) \Rightarrow e^2 = \frac{1}{2}$$
The product of focal distances from a point $P(x,y)$ on the ellipse is given by:
$$SP \cdot S’P = a^2 – e^2 x^2$$
The maximum value occurs when $x^2$ is minimum ($x=0$, at the co-vertices):
$$\max(SP \cdot S’P) = a^2 – 0 = 18$$
The minimum value occurs when $x^2$ is maximum ($x^2=a^2$, at the vertices):
$$\min(SP \cdot S’P) = a^2 – e^2 a^2 = a^2(1-e^2) = b^2 = 9$$
Calculate the sum:
$$\text{Sum} = 18 + 9 = 27$$
Ans. (4)
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