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 $\left(0, \frac{11}{6}\right)$