JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 2)
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the
eccentricity of the hyperbola is :
$${{13} \over 6}$$
2
$${{13} \over 12}$$
$${{13} \over 8}$$
Explanation
2b = 5 and 2ae = 13
b2 = a2(e2 $$-$$ 1) $$ \Rightarrow $$ $${{25} \over 4}$$ = $${{169} \over 4}$$ $$-$$ a2
$$ \Rightarrow $$ a $$=$$ 6 $$ \Rightarrow $$ e $$=$$ $${{13} \over {12}}$$
b2 = a2(e2 $$-$$ 1) $$ \Rightarrow $$ $${{25} \over 4}$$ = $${{169} \over 4}$$ $$-$$ a2
$$ \Rightarrow $$ a $$=$$ 6 $$ \Rightarrow $$ e $$=$$ $${{13} \over {12}}$$
Comments (0)
