Q.

Let the product of focal distances of the point P(4,23) on the hyperbola H : x2a2–y2b2=1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2+q2 is equal to __________.          [2025]


Ans.

(120)

We have, P(4,23)

Now, PS1·PS2=32          ... (i)

where S1=(ae,0) and S2=(–ae,0)

ALso, |PS1–PS2|=2a

Now, P(4,23) lies on H

∴  16a2–12b2=1

⇒ 16b2–12a2=a2b2          ... (ii)

Now, |PS1–PS2|2=4a2

⇒ PS12+PS22–2PS1·PS2=4a2

⇒ (ae–4)2+12+(ae+4)2+12–64=4a2

⇒ 2a2e2–8=4a2 ⇒ a2+b2–4=2a2

⇒ b2–a2=4          ... (iii)

From equation (ii) and (iii)

⇒ 16(a2+4)–12a2=a2(a2+4)

⇒ 16a2+64–12a2=a4+4a2

⇒ a4=64 ⇒ a2=8     ∴ b2=12

Now, p2+q2=4b2+4b4a2=120.