Question

The circle $x^2+y^2+6x-24y+72=0$ and hyperbola $x^2-y^2+6x+16y=46$ intersect at four distinct points. If these four points of intersection lie on a parabola, then

• A. the coordinates of the focus of the parabola is $(-3,2)$
• B. the coordinates of the focus of the parabola is $(-3,0)$
• C. the equation of the directrix of the parabola is $y=0$
• D. the length of the latus rectum of the parabola is $4$.

Answer 1

Answer: (A), (C), (D)

the length of the latus rectum of the parabola is $4$.

Let $S_1:x^2+y^2+6x-24y+72=0$
$S_2:x^2-y^2+6x+16y-46=0$
The family of curves will be,
$S_1+\lambda S_2=0\Rightarrow (1+\lambda )x^2+(1-\lambda )y^2+6(1+\lambda )x+(16\lambda -24)y+72-46\lambda =0$
Comparing it to general two degree curve,
$a{x}^2+b{y}^2+2hxy+2fx+2gy+c=0$

For the equation to represent parabola,
$h^2=ab$
$\Rightarrow (1+\lambda )(1-\lambda )=0\Rightarrow \lambda =\pm 1$

When $\lambda =-1$
$S_1-S_2=0\Rightarrow y^2-20y+59=0$
This is not a parabola.

When $\lambda =1$
$S_1+S_2=0\Rightarrow x^2+6x-4y+13=0\Rightarrow (x+3{)}^2=4(y-1)$
The focus is $(-3,1+1)=(-3,2)$
Directrix is $y=1-1=0$
Length of latus rectum is $4$