Question

A parabola is drawn with its vertex at $(0,-3)$, the axis of symmetry along the conjugate axis of the hyperbola $\frac{x^2}{49}-\frac{y^2}{9}=1$ and passing through the two foci of the hyperbola. The co-ordinates of the focus of the parabola are

• A. $\left(0,\frac{11}{6}\right)$
• B. $\left(0,-\frac{11}{6}\right)$
• C. $\left(0,\frac{11}{12}\right)$
• D. $\left(0,-\frac{11}{12}\right)$

Answer 1

Answer: (A)

Eqn. of hyperbola is $\frac{x^2}{49}-\frac{y^2}{9}=1$
Its conjugate axis is y-axis.
Also $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{49}}=\frac{\sqrt{58}}{7}$
$\therefore$ Foci of hyperbola $(\pm ae,0)$,
i.e. $(\pm \sqrt{58},0)$. Now equation of parabola with vertex at $(0,-3)$ and axis
along y-axis is $x^2=\ell (y+3)$
It passes through $(\pm \sqrt{58},0)\therefore 58=\ell (0+3)\Rightarrow \ell =\frac{58}{3}$
$\therefore$ Parabola is $x^2=\frac{58}{3}(y+3)$
Its focus is $\left(0,-3+\frac{58}{4.3}\right)$ or $\left(0,\frac{11}{6}\right)$