If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{6}{5}$ and $3 x+4 y-6=0$ is a focal chord of the hyperbola, then the length of transverse axis is equal to

JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

If the eccentricity of the hyperbola

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

\frac{6}{5}

and

3 x+4 y-6=0

is a focal chord of the hyperbola, then the length of transverse axis is equal to

View answer and explanation

Correct option: B

\frac{10}{3}

Option B (

\frac{10}{3}

) is correct because the length of the transverse axis is calculated using the eccentricity and the fact that the focal chord passes through a focus.

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\frac{x^2}{144}+\frac{y^2}{81}=1

m

\mathrm{M}

\mathrm{mM}

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\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

S

S'.

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PSS',

A

A

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(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

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

r

\frac{x^2}{25}+\frac{y^2}{16}=1

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O