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Hyperbola

Section: Mathematics

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Question : 15 of 98

Marks: +1, -0

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form

2 a

2 b

, respectively, and one focus and the corresponding directrix of this hyperbola be

(-5,0)

5 x+9=0

, respectively. If the product of the focal distances of a point

(\alpha, 2\sqrt{5})

on the hyperbola is

p

4 p

\_\_\_\_

[7 Apr 2025 Shift 2]

Your Answer:

Solution:

Equation of hyperbola is

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

Directrix:

x=\;\frac{-9}{5}

and corresponding foci

(-5,0)

\;\Rightarrow\;\; -\;\frac{a}{e}=-\;\frac{9}{5}\;\text{ and }\;-a e=-5

\;\Rightarrow\;\;\;\frac{9 e^2}{5}=5 \Rightarrow e=\sqrt{\;\frac{25}{9}}=\;\frac{5}{3}\Rightarrow a=3

\;\therefore\;\;b^2=a^2(e^2-1)=9\left(\;\frac{25}{9}-1\right)=16

Hyperbola

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

(\alpha, 2\sqrt{5})

lie on it

\Rightarrow\;\frac{\alpha^2}{9}-\;\frac{20}{16}=1\Rightarrow\alpha^2=\;\frac{36}{16}\times 9=\;\frac{81}{4}

Product for distance of (

x_1 y_1

) from the two foci

\;= (e x_1+a)\left| e x_1-a\right|

\;= e^2 x_1^2-a^2

\;\;\text{ For }\;(\alpha, 2\sqrt{5})\Rightarrow P=\;\frac{25}{9}\cdot\;\frac{81}{4}-9=\;\frac{189}{4}

\;4 P=189

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