If S and S' are the foci of the ellipse x^2/18 + y^2/9 = 1 and P be a point on the ellipse, then min(SP * S'P) + max(SP * S'P) is equal to :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2025

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Home Maths Conic SectionsQuestion #129

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 (S P \cdot S^{'} P) + max (S P \cdot S^{'} P)

is equal to :

Clue 1 of 3: Ellipse Properties

Recall the standard equation of an ellipse and the definitions of its semi-major axis, semi-minor axis, and eccentricity.

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Ninja Strategy•Estimate and Check Bounds

Calculate the minimum and maximum possible values for the product $S P \cdot S^{'} P$ . The minimum is 9 and the maximum is 18. The sum must be $9 + 18 = 27$ . Eliminate options that are too small or represent only one of the values.

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● Video Walkthrough

Step 1: Identify Ellipse Parameters✦ Active

From the given ellipse equation $\frac{x^{2}}{18} + \frac{y^{2}}{9} = 1$ , we identify the semi-major axis squared $a^{2} = 18$ and semi-minor axis squared $b^{2} = 9$ . Thus, $a = \sqrt{18} = 3 \sqrt{2}$ and $b = \sqrt{9} = 3$ . The eccentricity $e$ is calculated using $b^{2} = a^{2} (1 - e^{2})$ .

9 = 18 (1 - e^{2}) \Rightarrow \frac{1}{2} = 1 - e^{2} \Rightarrow e^{2} = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}

Step 2: Express Product of Focal Distances○ Expand

For any point $P (x, y)$ on the ellipse, the distances from the foci $S$ and $S^{'}$ are $S P = a - e x$ and $S^{'} P = a + e x$ (or vice versa). The product $S P \cdot S^{'} P$ is given by $a^{2} - e^{2} x^{2}$ . Substitute the values of $a^{2}$ and $e^{2}$ .

S P \cdot S^{'} P = 18 - \frac{1}{2} x^{2}

Step 3: Calculate Min and Max Values and Their Sum○ Expand

For a point on the ellipse, the $x$ -coordinate ranges from $- a$ to $a$ , i.e., $x \in [- 3 \sqrt{2}, 3 \sqrt{2}]$ . This means $x^{2} \in [0, (3 \sqrt{2})^{2}] = [0, 18]$ . Since $S P \cdot S^{'} P = 18 - \frac{1}{2} x^{2}$ is a decreasing function of $x^{2}$ :

min (S P \cdot S^{'} P) occurs when x^{2} = 18 : 18 - \frac{1}{2} (18) = 18 - 9 = 9

max (S P \cdot S^{'} P) occurs when x^{2} = 0 : 18 - \frac{1}{2} (0) = 18

The required sum is $min (S P \cdot S^{'} P) + max (S P \cdot S^{'} P) = 9 + 18 = 27$ .

💡 Teacher's Secret Hint

The minimum product occurs at the vertices on the major axis, and the maximum product occurs at the co-vertices on the minor axis.

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Maths Question 4 – JEE-MAIN 2025

Clue 1 of 3: Ellipse Properties

🧠 Full Solution Path