Let the lengths of the transverse and conugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (–5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances

Question

Let the lengths of the transverse and conugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (–5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point $(α, 2 \sqrt{5})$ on the hyperbola is p, then 4p is equal to __________. [2025]

Smart Achievers · Accepted Answer

(189)

Given, focus = (–5, 0)

$\Rightarrow a e = 5 and \frac{a}{e} = \frac{9}{5}$

Also, $a = 3, e = \frac{5}{3} and b = 4$

$\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Since, hyperbola passes through $(α, 2 \sqrt{5})$ , we get

$\frac{α^{2}}{9} - \frac{4 (5)}{16} = 1$

$\Rightarrow α^{2} = 9 (\frac{36}{16}) = \frac{81}{4}$

Now, $P = e^{2} α^{2} - a^{2} = \frac{25}{9} \times \frac{81}{4} - 9 = \frac{189}{4}$

$∴ 4 p = 189$ .

Solution

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Solution

Ans. (189) Given, focus = (–5, 0) ⇒ ae=5 and ae=95 Also, a=3, e=53 and b=4 ⇒x29–y216=1 Since, hyperbola passes through (α,25), we get α29–4(5)16=1 ⇒ α2=9(3616)=814 Now, P=e2α2–a2=259×814–9=1894 ∴ 4p=189.

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