Question

Find the equation of parabola whose directrix is 2x - 3y + 4 = 0 and vertex at (5-4).​

Answer 1

The required equation is $9x^2+4y^+12xy2-114x+128y+12y+517+12xy=0$

Step-by-step explanation:

Given:

Let S(5,-4) and equation of the directrix is 2x - 3y + 4 = 0.

Let P is a point whose coordinates are (x,y) which moves such that, where PM is the length of the perpendicular drawn from P to  directrix.

$(x - 5)^2+(y - (-4))^2 =[\frac{|2x - 3y + 4 |}{\sqrt{4+9} }]^2$

$(x - 5)^2+(y +4)^2 =[\frac{|2x - 3y + 4 |}{\sqrt{4+9} }]^2$

$x^2-10x+25+y^2+8y+16=[\frac{|2x - 3y + 4|}{\sqrt{13} }]^2$

$x^2-10x+25+y^2+8y+16=\frac{1}{13}[|2x - 3y + 4|]^2$

$x^2-10x+25+y^2+8y+16=\frac{1}{13}[|(2x - 3y) + (4)|]^2$

$x^2-10x+25+y^2+8y+16=\frac{1}{13}[(2x-3y)^2- 2\times 4(2x-3y)+4^2]$   $[(a+b)^2=a^2+2ab+b^2]$

$13[x^2-10x+25+y^2+8y+16]=4x^2-12xy+9y^2- 8(2x-3y)+16$

$13[x^2-10x+y^2+8y+41]=4x^2-12xy+9y^2- 16x-24y+16$

$13x^2-130x+13y^2+104y+533=4x^2-12xy+9y^2- 16x-24y+16$

$13x^2-4x^2+13y^2-9y^2-130x+16x+104y+24y+533-16+12xy=0$

$9x^2+4y^2+12xy-114x+128y+12y+517=0$.

Hence this the required equation.