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The equation of the hyperbola, in standard form, whose vertices are at ( \pm 5,0) and foci at ( \pm 7,0), is

Option: 1

\mathrm{\frac{x^2}{25}-\frac{y^2}{24}=1}


Option: 2

\mathrm{\frac{x^2}{24}-\frac{y^2}{25}=1}


Option: 3

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


Option: 4

\mathrm{\text { none of these }}


Answers (1)

best_answer

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

the vertex A is \mathrm{(a, 0)} and the focus S is \mathrm{(a e, 0).}

Here, A is \mathrm{(5,0)} and S is \mathrm{(7,0)}
\mathrm{\Rightarrow \quad a=5 \, \, and\, \, a e=7 \Rightarrow e=\frac{7}{5}}

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

           \mathrm{ \begin{aligned} b^2 & =25\left(\frac{49}{25}-1\right) \\ & =24 \end{aligned} }

\therefore  The hyperbola is \mathrm{\frac{x^2}{25}-\frac{y^2}{24}=1}

The answer is (a) 

Posted by

shivangi.bhatnagar

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