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 coordinates of the focus of the parabola are :

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

Answer 1

The correct option is A $(0,\frac{11}{6})$

Equation 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 is $(\pm ae,0)\Rightarrow (\pm \sqrt{58},0)$

Now equation of parabola with vertex at $(0,-3)$ and axis along y-axis is $x^2=l(y+3)$

It passes through $(\pm \sqrt{58},0)$

$\therefore 58=l(0+3)\Rightarrow l=\frac{58}{3}$

$\therefore$ parabola is $x^2=\frac{58}{3}(y+3)$

Its focus is $(0,-3+\frac{58}{4.3})\equiv (0,\frac{11}{6})$