# Thread: Use the definition of a parabola and the distance

1. ## Use the definition of a parabola and the distance

Hi all

Use the definition of a parabola and the distance formula to find the equation of a parabola with

a ) directix x = -4 and focus (2,2 )
B ) directix x = 2 and focus (6,-4 )

How i solve like this queation please hle me the steps to solve that

thanks

2. For any point on parabola

its distance from directrix and its distance from focus are equal

So

Distance of a point(h,k) from a line x= a

is given by

$L= |{h-a}|$

Distance between two points (h,k) and (x,y) is given by

$D= \sqrt{(h-x)^2+(k-y)^2}$

D= L

So $|{h-a}| = \sqrt{(h-x)^2+(k-y)^2}$

You have been given all the values
Solve it and ask if you still find it difficult.

3. Originally Posted by r-soy
Hi all

Use the definition of a parabola and the distance formula to find the equation of a parabola with

a ) directix x = -4 and focus (2,2 )
B ) directix x = 2 and focus (6,-4 )

How i solve like this queation please hle me the steps to solve that

thanks

Definition: a (plane) parabola is the locus of all points in the plane which are equidistant from a given line, called directrix, and a given fixed point called Focus.

So , for example, in (a) you're looking for all the points $(x,y)\in\mathbb{R}^2$ s. t. their distances from the line $x+4=0$ and from the point $(2,2)$ are the same, i.e.:

$\frac{|x+4|}{1}=\sqrt{(x-2)^2+(y-2)^2}$ ....well, now solve this equation (strongly recommended to square both sides) and find out the equation.

Tonio

4. I try to solve

(x+4)^2 = ( X-2)^2 + (y-2) ^2

x^2-16x+16=X^2-4x+4 + y2-4y+4

x^2-16x+16-X^2+4x-4 -y2+4y-4
12x-y^2-8 = 0

5. Originally Posted by r-soy
I try to solve

(x+4)^2 = ( X-2)^2 + (y-2) ^2

x^2-16x+16=X^2-4x+4 + y2-4y+4

It must be $+16x$ on the left side above, but basically it is correct.

Toni

x^2-16x+16-X^2+4x-4 -y2+4y-4
12x-y^2-8 = 0