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Ellipse question

2025 · 2 Apr · Shift 1 · Q30

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2025 · 2 Apr · Shift 1 · Q30

Ellipse question

2025 · 2 Apr · Shift 1 · Q30

JEE MainMathematicsEllipseMCQ+4 / −1

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 \cdot S^{\prime} P\right)+\max \left(S P \cdot S^{\prime} P\right)

is equal to :

A
$3(6+\sqrt{2})$
B
$3(1+\sqrt{2})$
C
27
D
9

View written solutionFree

Correct answer: C

Write the ellipse in standard form

Given

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

So,

with major axis along the $x$ -axis.

Hence,

c^2=a^2-b^2=18-9=9 \implies c=3

Therefore the foci are

S=(3,0),\qquad S'=(-3,0).

Use the focal distance property

For any point $P$ on the ellipse, if

then

We need the minimum and maximum of

Parametrize the point on the ellipse

Let $P=(a\cos\theta,b\sin\theta)=(3\sqrt2\cos\theta,3\sin\theta).$

Now,

But for an ellipse, the standard formulas for focal distances are

if foci are taken symmetrically; equivalently, using parameter form,

where

Thus,

Since

we get

Find minimum and maximum

Because

we have

Maximum occurs when $\cos^2\theta=0$ : $\max(r_1r_2)=18.$
Minimum occurs when $\cos^2\theta=1$ : $\min(r_1r_2)=9.$

Therefore,

\min(SP\cdot S'P)+\max(SP\cdot S'P)=9+18=27.

Check options

The value is $27,$ which matches Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.

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