The equation of hyperbola whose eccentricity is 53 and distance between the foci is 10 units is

e=53 2ae=10⇒a=3 b2=a2(e2−12)=9(−1+259)=16 x29−y216=1⇒16x2−9y2=144

The equation of hyperbola whose eccentricity is 53 and distance between the foci is 10 units is

e=53 2ae=10⇒a=3 b2=a2(e2−12)=9(−1+259)=16 x29−y216=1⇒16x2−9y2=144

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

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$16 x^{2} - 9 y^{2} = 16$

$16 x^{2} - 9 y^{2} = 9$

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

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

answer is D.

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Detailed Solution

$e = \frac{5}{3} 2 a e = 10 \Rightarrow a = 3 b^{2} = a^{2} (e^{2} - 1^{2}) = 9 (- 1 + \frac{25}{9}) = 16 \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 \Rightarrow 16 x^{2} - 9 y^{2} = 144$

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