Question

Find the coordinates of a point on the parabola $y^2=8x$ whose focal distance is $4$.

Answer 1

Given: parabola $y^2=8x$ …(i)

And standard parabola : $y^2=4ax$ … (ii).

Compare both equation (i) and (ii), we get

$a=2$

Now Focus of parabola is : $F(2,0)$

$[\because Focusforstandardparabola{y}^2=4axis:(a,0\left)\right]$

Let $P(\alpha ,\beta )$ is any point on parabola : $y^2=8x$

Given focal distance is $4$

$\Rightarrow PF=4$

Distance between points $(x_1,y_1)$ and $(x_2,y_2)$

$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\Rightarrow$ Distance between $F(2,0)$ and point $P(\alpha ,\beta )$:

$\sqrt{(\alpha -2{)}^2+(\beta -0{)}^2}=4$

Squaring on both sides

$\Rightarrow (\alpha -2{)}^2+{\beta }^2=16$ …(iii)

The point $P(\alpha ,\beta )$ lies on parabola $y^2=8x$

$\Rightarrow {\beta }^2=8\alpha$ …(iv)

Now using equation (iii) and (iv).

$\Rightarrow (\alpha -2{)}^2+8\alpha =16$

$\Rightarrow {\alpha }^2+4\alpha +4=16$

$\Rightarrow (\alpha +2)=\pm 4$

$\Rightarrow \alpha =-6,2$

$\alpha =-6$ is not possible because $x$ coordinate of any point lying on given parabola $(y^2=8x)$ is always positive.

Putting $\alpha =2$ in equation (iv)

$\Rightarrow \beta =\pm 4$

$\therefore$ The points are $(2,4)$ and $(2,-4)$

Hence, points on parabola are: $(2,4)$ and $(2,-4)$.