Let the sum of the focal distances of the point P(4, 3) on the hyperbola H : x^2/a^2 - y^2/b^2 = 1 be 8sqrt(5/3). If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then 9l^2 + 6m is equal to :

The correct answer is Option 2. 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 #142

Let the sum of the focal distances of the point

P (4, 3)

on the hyperbola

H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

8 \sqrt{\frac{5}{3}}

. If for

H

, the length of the latus rectum is

l

and the product of the focal distances of the point

P

m

, then

9 l^{2} + 6 m

is equal to :

Clue 1 of 3: Hyperbola Properties

Recall the definitions of focal distances, eccentricity, and latus rectum for a hyperbola.

🧠 Full Solution Path

Step 1: Determine Eccentricity and Hyperbola Parameters✦ Active

For a point $P (x, y)$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the sum of focal distances is $2 e x$ . Given $P (4, 3)$ and the sum is $8 \sqrt{\frac{5}{3}}$ , we have $2 e (4) = 8 \sqrt{\frac{5}{3}}$ , which implies $e = \sqrt{\frac{5}{3}}$ . Thus, $e^{2} = \frac{5}{3}$ . The relation $b^{2} = a^{2} (e^{2} - 1)$ gives $b^{2} = a^{2} (\frac{5}{3} - 1) = \frac{2 a^{2}}{3}$ .

2 e x = 8 \sqrt{\frac{5}{3}} ⟹ 2 e (4) = 8 \sqrt{\frac{5}{3}} ⟹ e = \sqrt{\frac{5}{3}}

b^{2} = a^{2} (e^{2} - 1) = a^{2} (\frac{5}{3} - 1) = \frac{2 a^{2}}{3}

Step 2: Calculate

a^{2}

and

b^{2}

○ Expand

Since $P (4, 3)$ lies on the hyperbola, substitute its coordinates into the equation: $\frac{4^{2}}{a^{2}} - \frac{3^{2}}{b^{2}} = 1$ . Substitute $b^{2} = \frac{2 a^{2}}{3}$ into this equation to solve for $a^{2}$ and then $b^{2}$ .

\frac{16}{a^{2}} - \frac{9}{b^{2}} = 1 ⟹ \frac{16}{a^{2}} - \frac{9}{\frac{2 a^{2}}{3}} = 1

\frac{16}{a^{2}} - \frac{27}{2 a^{2}} = 1 ⟹ \frac{32 - 27}{2 a^{2}} = 1 ⟹ \frac{5}{2 a^{2}} = 1

a^{2} = \frac{5}{2}

b^{2} = \frac{2}{3} a^{2} = \frac{2}{3} \cdot \frac{5}{2} = \frac{5}{3}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when substituting $b^{2}$ to avoid errors.

Step 3: Calculate

l

m

, and the final expression○ Expand

The length of the latus rectum is $l = \frac{2 b^{2}}{a}$ . The product of focal distances is $m = e^{2} x^{2} - a^{2}$ . Substitute the values of $a^{2}, b^{2}, e^{2}, x$ to find $l^{2}$ and $m$ , then calculate $9 l^{2} + 6 m$ .

l = \frac{2 b^{2}}{a} = \frac{2 (5 / 3)}{\sqrt{5 / 2}} = \frac{10 / 3}{\sqrt{5} / \sqrt{2}} = \frac{10 \sqrt{2}}{3 \sqrt{5}} = \frac{10 \sqrt{10}}{15} = \frac{2 \sqrt{10}}{3}

l^{2} = {(\frac{2 \sqrt{10}}{3})}^{2} = \frac{4 \cdot 10}{9} = \frac{40}{9}

m = e^{2} x^{2} - a^{2} = (\frac{5}{3}) (4^{2}) - \frac{5}{2} = \frac{5}{3} (16) - \frac{5}{2} = \frac{80}{3} - \frac{5}{2}

m = \frac{160 - 15}{6} = \frac{145}{6}

9 l^{2} + 6 m = 9 (\frac{40}{9}) + 6 (\frac{145}{6}) = 40 + 145 = 185

💡 Teacher's Secret Hint

Double-check calculations for $l^{2}$ and $m$ before the final substitution.

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Hyperbola Properties

🧠 Full Solution Path