Question

Write the equation of the parabola whose vertex is at (-3,0) and the directrix is x+5=0

Answer 1

Here, vertex = (-3,0)

$\therefore a=-3anddirectrix,x+5=0$

Since, axis of the parabola is a line perpendicular to directrix and A is the mid-point of AS.

$Then,-3=\frac{x_1-5}{2}$

$\Rightarrow -6=x_1-5\Rightarrow x_1=-1$

$0=\frac{0+y_1}{2}\Rightarrow =0$

$\therefore S=(-1,0)$

$\therefore PM=PS$

$\Rightarrow |x+5|=\sqrt{(x+1{)}^2+y^2}$

$\Rightarrow x^2+2x+1+y^2=x^2+10x+25$

$\Rightarrow y^2=8x+24$

$\Rightarrow y^2=8(x+3)$