Let the product of the focal distances of the point P(4,2√3) on the hyperbola H: x^2/a^2 - y^2/b^2 = 1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p^2 + q^2 is equal to _______

The correct answer is Option 120. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q23 – JEE-MAIN 2025

← Back

Home Maths Conic SectionsQuestion #136

Let the product of the focal distances of the point

P (4, 2 \sqrt{3})

on the hyperbola

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

be 32. Let the length of the conjugate axis of H be

p

and the length of its latus rectum be

q

. Then

p^{2} + q^{2}

is equal to _______

Enter Calculated Numerical Value:

Clue 1 of 3: Understand Hyperbola Properties

Recall the definitions and formulas for focal distances, conjugate axis, and latus rectum of a hyperbola.

🧠 Full Solution Path

Step 1: Formulate Equations from Given Information✦ Active

The product of focal distances for a point $(x_{1}, y_{1})$ on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e^{2} x_{1}^{2} - a^{2}$ . Given $P (4, 2 \sqrt{3})$ , we have $e^{2} (4^{2}) - a^{2} = 32$ , which simplifies to $16 e^{2} - a^{2} = 32$ (Equation 1).

The point $P (4, 2 \sqrt{3})$ lies on the hyperbola, so $\frac{4^{2}}{a^{2}} - \frac{(2 \sqrt{3})^{2}}{b^{2}} = 1$ , which simplifies to $\frac{16}{a^{2}} - \frac{12}{b^{2}} = 1$ (Equation 2).

The relationship between $a, b, e$ for a hyperbola is $b^{2} = a^{2} (e^{2} - 1)$ , which can be rewritten as $e^{2} = 1 + \frac{b^{2}}{a^{2}}$ (Equation 3).

Step 2: Solve for

a^{2}

and

b^{2}

○ Expand

Substitute Equation 3 into Equation 1: $16 (1 + \frac{b^{2}}{a^{2}}) - a^{2} = 32$ . This simplifies to $16 + \frac{16 b^{2}}{a^{2}} - a^{2} = 32$ , or $\frac{16 b^{2}}{a^{2}} - a^{2} = 16$ (Equation 4).

From Equation 2, we can express $b^{2}$ in terms of $a^{2}$ : $\frac{12}{b^{2}} = \frac{16}{a^{2}} - 1 = \frac{16 - a^{2}}{a^{2}}$ , so $b^{2} = \frac{12 a^{2}}{16 - a^{2}}$ (Equation 5).

Substitute Equation 5 into Equation 4: $\frac{16}{a^{2}} (\frac{12 a^{2}}{16 - a^{2}}) - a^{2} = 16$ . This simplifies to $\frac{192}{16 - a^{2}} - a^{2} = 16$ . Multiplying by $(16 - a^{2})$ gives $192 - a^{2} (16 - a^{2}) = 16 (16 - a^{2})$ . Expanding and rearranging: $192 - 16 a^{2} + a^{4} = 256 - 16 a^{2}$ . This leads to $a^{4} = 64$ , so $a^{2} = 8$ .

Substitute $a^{2} = 8$ back into Equation 5 to find $b^{2}$ : $b^{2} = \frac{12 \times 8}{16 - 8} = \frac{96}{8} = 12$ .

Step 3: Calculate

p^{2} + q^{2}

○ Expand

The length of the conjugate axis is $p = 2 b$ . So, $p^{2} = (2 b)^{2} = 4 b^{2} = 4 \times 12 = 48$ .

The length of the latus rectum is $q = \frac{2 b^{2}}{a}$ . So, $q^{2} = {(\frac{2 b^{2}}{a})}^{2} = \frac{4 b^{4}}{a^{2}} = \frac{4 \times (12)^{2}}{8} = \frac{4 \times 144}{8} = \frac{576}{8} = 72$ .

Finally, $p^{2} + q^{2} = 48 + 72 = 120$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 23 – JEE-MAIN 2025

Clue 1 of 3: Understand Hyperbola Properties

🧠 Full Solution Path