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NDA Math Conic Section

Question : 30 of 34

Marks: +1, -0

The hyperbola

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

(3\sqrt{5},1)

\frac{4}{3}

Solution:

Consider the given equation of the hyperbola and the point lying on it.The length of the latus rectum will be

\frac{2b^2}{a}=\frac{4}{3}

\therefore b^2= \frac{2}{3}a

Now, considering the equation of the hyperbola and the point lying on it,we get

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

P(3\sqrt{5},1)

Substituting the given point in the equation of the hyperbola, we get

\frac{45}{a^2}-\frac{1}{b^2}=1

\frac{45}{a^2}-\frac{3}{2a}=1

90-3a=2a^2

2a^2+3a-90=0

a=6

(taking positive value)

\therefore b= \sqrt{\frac{2a}{3}}

=\sqrt{\frac{2\times6}{3}}=2

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

\Rightarrow \frac{x^2}{6^2}-\frac{y^2}{2^2}=1

Conjugate axis length

=2b=2(2)=4

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