Question

The auxiliary circle of $\frac{x^2}{9}+\frac{y^2}{4}=1$ touches a parabola having axis of symmetry as $y-$axis at its vertex . If the latus rectum of parabola is equal to radius of director circle of auxiliary circle, then the equation of parabola can be

• A. $x^2=\sqrt{18}(y-3)$
• B. $x^2=\sqrt{16}(y+3)$
• C. $x^2=-\sqrt{16}(y-3)$
• D. $x^2=-\sqrt{18}(y+3)$

Answer 1

Answer: (A), (D)

$x^2=-\sqrt{18}(y+3)$

Auxiliary circle of ellipse is $x^2+y^2=9$
Director circle equatio for auxilary circle is $x^2+y^2=18\Rightarrow (r=\sqrt{18})$
So, length of latus rectum $=\sqrt{18}$ and vertex of parabola is $(\pm 3,0)$ as parabola touches the circle at vertex only
So. equation of parabola is $(x-0{)}^2=\pm \sqrt{18}(y\pm 3)$
therefore equation of upward parabola will be $x^2=\sqrt{18}(y-3)$
And, equation of downward parabola will be $x^2=-\sqrt{18}(y+3)$