The vertices of the hyperbola $7 x^2-49 y^2=343$ having eccentricity ' $4 / 3^{\prime}$ is
The vertices of the hyperbola $7 x^2-49 y^2=343$ having eccentricity ' $4 / 3^{\prime}$ is
$(0,0)$
$( \pm 3,0)$
$(0, \pm 5)$
$( \pm 7,0)$
Solution
Equation of given hyperbola is
$
7 x^2-49 y^2=343 \Rightarrow \frac{x^2}{49}-\frac{y^2}{7}=1
$
So, the coordinate of vertices are $( \pm 7,0)$. Hence, option (4) is correct