A line passing through the point P(√5, √5) intersects the ellipse x^2/36 + y^2/25 = 1 at A and B such that (PA)(PB) is maximum. Then 5(PA^2 + PB^2) is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q12 – JEE-MAIN 2025

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Home Maths Conic SectionsQuestion #39

A line passing through the point

P (\sqrt{5}, \sqrt{5})

intersects the ellipse

\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1

A

and

B

such that

(P A) (P B)

is maximum. Then

5 (P A^{2} + P B^{2})

is equal to :

Clue 1 of 3: Parametric Equation of a Line

Represent the line passing through $P (x_{1}, y_{1})$ in parametric form $x = x_{1} + r \cos θ$ , $y = y_{1} + r \sin θ$ .

🧠 Full Solution Path

Step 1: Formulate Quadratic in 'r' and Maximize PA.PB✦ Active

Let the line passing through $P (x_{1}, y_{1}) = (\sqrt{5}, \sqrt{5})$ be $x = x_{1} + r \cos θ$ and $y = y_{1} + r \sin θ$ . Substituting this into the ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (where $a^{2} = 36, b^{2} = 25$ ) gives a quadratic equation in $r$ : $A r^{2} + B r + C = 0$ . The roots $r_{1}, r_{2}$ represent $P A$ and $P B$ . The product $P A \cdot P B = | r_{1} r_{2} | = | \frac{C}{A} |$ . Here, $C = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1 = \frac{5}{36} + \frac{5}{25} - 1 = \frac{5}{36} + \frac{1}{5} - 1 = \frac{25 + 36 - 180}{180} = - \frac{119}{180}$ . And $A = \frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}$ . Thus, $P A \cdot P B = \frac{119 / 180}{\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25}}$ . To maximize $P A \cdot P B$ , we must minimize the denominator $f (θ) = \frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25} = \frac{1 - \sin^{2} θ}{36} + \frac{\sin^{2} θ}{25} = \frac{1}{36} + \sin^{2} θ (\frac{1}{25} - \frac{1}{36}) = \frac{1}{36} + \frac{11}{900} \sin^{2} θ$ . This is minimized when $\sin^{2} θ = 0$ , which means $θ = 0$ (horizontal line). The minimum value of the denominator is $\frac{1}{36}$ . Therefore, $(P A \cdot P B)_{m a x} = \frac{119 / 180}{1 / 36} = \frac{119}{5}$ .

Step 2: Determine the Line Equation and Intersection Points○ Expand

Since $\sin^{2} θ = 0$ , the line is horizontal, passing through $P (\sqrt{5}, \sqrt{5})$ , so its equation is $y = \sqrt{5}$ . Substitute $y = \sqrt{5}$ into the ellipse equation: $\frac{x^{2}}{36} + \frac{(\sqrt{5})^{2}}{25} = 1 ⟹ \frac{x^{2}}{36} + \frac{5}{25} = 1 ⟹ \frac{x^{2}}{36} + \frac{1}{5} = 1 ⟹ \frac{x^{2}}{36} = \frac{4}{5} ⟹ x^{2} = \frac{144}{5}$ . So $x = \pm \frac{12}{\sqrt{5}}$ . The intersection points are $A = (\frac{12}{\sqrt{5}}, \sqrt{5})$ and $B = (- \frac{12}{\sqrt{5}}, \sqrt{5})$ . The point $P$ is $(\sqrt{5}, \sqrt{5})$ .

Step 3: Calculate

5 (P A^{2} + P B^{2})

○ Expand

The distances $P A$ and $P B$ are horizontal distances from $P$ to $A$ and $B$ . $P A = | x_{A} - x_{P} | = | \frac{12}{\sqrt{5}} - \sqrt{5} | = | \frac{12 - 5}{\sqrt{5}} | = \frac{7}{\sqrt{5}}$ . $P B = | x_{B} - x_{P} | = | - \frac{12}{\sqrt{5}} - \sqrt{5} | = | - \frac{12 + 5}{\sqrt{5}} | = \frac{17}{\sqrt{5}}$ . Now, calculate $P A^{2} + P B^{2}$ : $P A^{2} = (\frac{7}{\sqrt{5}})^{2} = \frac{49}{5}$ . $P B^{2} = (\frac{17}{\sqrt{5}})^{2} = \frac{289}{5}$ . So, $P A^{2} + P B^{2} = \frac{49}{5} + \frac{289}{5} = \frac{338}{5}$ . Finally, $5 (P A^{2} + P B^{2}) = 5 \cdot \frac{338}{5} = 338$ .

💡 Teacher's Secret Hint

Ensure to use the correct coordinates for P, A, and B when calculating distances.

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Parametric Equation of a Line

🧠 Full Solution Path