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  • The equation of a hyperbola in standard form whose length of the latus rectum is 8 and eccentricity is <img alt="\mathrm{\frac{3}{\sqrt{5}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

The equation of a hyperbola in standard form whose length of the latus rectum is 8 and eccentricity is \mathrm{\frac{3}{\sqrt{5}}} is

Option: 1

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


Option: 2

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


Option: 3

\mathrm{\frac{x^2}{25}-\frac{y^2}{20}=1}


Option: 4

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


Answers (1)

best_answer

Let the hyperbola be  \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}

Length of the latus rectum \mathrm{=\frac{2 b^2}{a}=8}

 \mathrm{\Rightarrow \quad b^2=4 a}                                        \mathrm{........(1)}

Also \mathrm{b^2=a^2\left(e^2-1\right)}

Or \mathrm{b^2=a^2\left(\frac{9}{5}-1\right)=\frac{4 a^2}{5}}             .........(2)

From (1) and (2),

\mathrm{ \therefore \quad \frac{4 a^2}{5}=4 a \text { or } a=5 }

Then, \mathrm{b^2=4 a=4 \times 5=20}

The required equation of the hyperbola is

                 \mathrm{ \frac{x^2}{25}-\frac{y^2}{20}=1 }

The answer is (c)

Posted by

Devendra Khairwa

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