Question

any point on the parabola whose focus is (0,1) and the directrix is x+2=0 is given by

Answer 1

Answer:

$(y-1)^2=4(x+1)$

Step-by-step explanation:

Focus F(0,1) and directrix D: x+2=0 of a parabola.

Let we take any point on parabola be P(h,k)

Distance of any point on parabola and focus is equal to distance of point and directrix.

Using distance formula, Find the distance of Focus (0,1) and point (h,k)

$FP=\sqrt{(h-0)^2+(k-1)^2}\Rightarrow \sqrt{h^2+k^2-2k+1}$

Distance of point (h,k) and line x+2=0

Using point line distance formula.

$DP=h+2$

$\sqrt{h^2+k^2-2k+1}=h+2$

$h^2+k^2-2k+1=(h+2)^2$

$h^2+k^2-2k+1=h^2+4+4h$

$k^2-2k+1-4-4h=0$

Replace h->x and k->y

$y^2-2y+1-4-4x=0$

$(y-1)^2=4(x+1)$

This is equation parabola.