For the hyperbola x^ 2 a^ 2 - y^ 2 b^ 2 = 1, distance between the foci is 10 ~units. From the point (2, 3 ), perpendicular tangents are drawn to the hyperbola, then the value of |(b)/(a)| is

Correct answer: C. 0. 75 Let,& e be the eccentricity of the hyperbola Now,~ 2ae = 10 ⇒ a^ 2 e^ 2 = 25 ⇒ a^ 2 + b^ 2 = 25 Also (2, 3 ) lies on the director circle~ x^ 2 + y^ 2 = a^ 2 - b^ 2 ⇒ 7 = a^ 2 - b^ 2 ⇒ a^ 2 = 16, b^ 2 = 9 ⇒ |(b)/(a)| = (3)/(4) = 0 . 75

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

For the hyperbola x^ 2 a^ 2 - y^ 2 b^ 2 = 1, distance between the...

For the hyperbola

\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1,

distance between the foci is

10

~units. From the point \left(2, \sqrt{3}\right),

perpendicular tangents are drawn to the hyperbola, then the value of \left|\dfrac{b}{a}\right|

Options

A $0. 25$
B $5$
C $0. 75$
D $1$

Correct answer

C. $0. 75$

Step-by-step solution

Let,&

e

be the eccentricity of the hyperbola
Now,~

2ae = 10 ⇒ a^{2}e^{2} = 25

⇒ a^{2} + b^{2} = 25

Also

\left(2, \sqrt{3}\right)

lies on the director circle~

x^{2} + y^{2} = a^{2} - b^{2}

⇒ 7 = a^{2} - b^{2} ⇒ a^{2} = 16, b^{2} = 9

⇒ \left|\dfrac{b}{a}\right| = \dfrac{3}{4} = 0 . 75

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