Question

find the equation of the parabola whose focus is (-1,2) and directrix is x-2y-15=0

Answer 1

using the definition,
√((x+1)²+(y-2)²)=|x-2y-15|/√(1+4)

squaring both sides
5*(x²+2x+1+y²-4y+4)=x²+4y²+225-4xy-30x+60y
4x²+y²+4xy+40x-80y-200=0

Answer 2

The equation of the parabola is 4x^2+y^2+4xy+40x-80y-240=0

Step-by-step explanation:

Given:

focus = (-1,2)

directrix is x-2y-15=0

Let P(x,y) be any point on the parabola whose focus S(–1,2) and directrix is x-2y-15=0

For a parabola

SP=PM

$SP^2=PM^2$

$(x-(-1))^2+(y-2)^2=[\frac{x-2y-15}{\sqrt{(-1)^2+(2)^2} }]^2$

$(x+1)^2+(y-2)^2=[\frac{x-2y-15}{\sqrt{5} }]^2$

$(x+1)^2+(y-2)^2=\frac{(x-2y-15)^2}{{5} }$

$5[(x^2+2x+1)+(y^2-4y-4)]=(x-2y-15)(x-2y-15)$

$5x^2+10x+5+5y^2-20y-20=x^2-2xy-15x-2xy+4y^2+30y-15x+30y+225$

$5x^2+5y^2+10x-20y-15=x^2+4y^2-30x-4xy+60y+225$

$5x^2-x^2+5y^2-4y^2+10x+30x-20y-60y-15-225+4xy=0$

$4x^2+y^2+4xy+40x-80y-240=0$

This is the required equation of the parabola.