Question

The circle $x^2+y^2+6x-24y+72=0$ and hyperbola $x^2-y^2+6x+16y-46=0$ intersect at four distinct points. These four points lie on a parabola. Then

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

Answer 1

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

length of latus rectum of parabola is $4$


Curve intersecting circle and hyperbola is $C+\lambda H=0$
$\Rightarrow (1+\lambda )x^2+(1-\lambda )y^2+6(1+\lambda )x-(24-16\lambda )y+72-46\lambda =0$
Given that above curve is a parabola
$\therefore h^2=ab$
$\Rightarrow 0=(1+\lambda )(1-\lambda )$
$\Rightarrow \lambda =-1,1$

For $\lambda =-1,$ we have
$2y^2-40y+118=0$
which is not a parabola as it represents two points on $y-$axis.

For $\lambda =1,$ we have
$2x^2+12x-8y+26=0$
$\Rightarrow x^2+6x-4y+13=0$
$\Rightarrow (x+3{)}^2=4(y-1)$
which represents a parabola.
Vertex $:(-3,1)$
Focus $:(-3,2)$
Equation of directrix $:y=0$
Length of latus rectum $=4\times 1=4$