The equation of the hyperbola whose eccentric | vedclass

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(D) Given,eccentricity $e = \frac{5}{3}$ and distance between foci $2ae = 10$. $2a(\frac{5}{3}) = 10 \Rightarrow a = 3$. For a hyperbola,$c^2 = a^2 +

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$16x^2 - 9y^2 = 144$

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(D) Given,eccentricity $e = \frac{5}{3}$ and distance between foci $2ae = 10$. $2a(\frac{5}{3}) = 10 \Rightarrow a = 3$. For a hyperbola,$c^2 = a^2 + b^2$,where $c = ae$. $(ae)^2 = a^2 + b^2 \Rightarrow 5^2 = 3^2 + b^2$. $25 = 9 + b^2 \Rightarrow b^2 = 16$. The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substituting the values,we get $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Multiplying by $144$,we get $16x^2 - 9y^2 = 144$.

The equation of the hyperbola whose eccentricity is $\frac{5}{3}$ and the distance between the foci is $10$ units is:

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