Question

Find the equation of parabola whose vertex is origin and directrix is x+y=2​

Answer 1

$\huge{\underline{\tt{\red{Answer}}}}\leadsto \boxed{(x+1)^2 +(y+1)^2 =\bigg(\dfrac{x+y-2}{\sqrt{1^2+1^2}}\bigg)^2}$

___________________________________

Given:

•Vertex=(0,0)

•directrix $\implies$ x+y-2=0

•AV=VB

___________________________________

$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

$\huge\star$ Foot Of Perpendicular = $\dfrac{x-x_1}{a} = \dfrac{y-y_2}{b} = \dfrac{-ax_1+by_1+c}{a^2+b^2}$

$\huge\star$ Equation of Parabola= $(x-\alpha)^2+(y-\beta)^2 =\bigg( \dfrac{lx+my+n}{\sqrt{l^2+m^2}}\bigg)^2$

___________________________________

$\huge{\underline{\tt{SoluTion}}}$

A= Foot of Perpendicular

$\dfrac{x-0}{1}=\dfrac{y-0}{1} = \dfrac{-(0+0-2)}{1^2+1^2} \\ \\ x=y=\dfrac{-(-2)}{2} \\ \\ x=y=1 \\ \\ \therefore Coordinates\:of\:BV=(x,y)=(-1,-1)$

So now Equation of Parabola

$\implies \boxed{(x+1)^2 +(y+1)^2 =\bigg(\dfrac{x+y-2}{\sqrt{1^2+1^2}}\bigg)^2}$

Answer 2

Answer:

Here is your answer ll

Hope it helps you