Question

The focal distance of a point on the parabola $y^2=8x$ is $4$; find the co-ordinates of the point.

Answer 1

Let the coordinate of that point is $P(x,y)$

Equation of parabola, $y^2=8x$ .............1

coordinate of focus $F(2,0)$

According to question

$PF=4$

$\sqrt{{(x-2)}^2+{(y-0)}^2}=4$

By squaring both side

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

from eq.1

$x^2+4-4x+8x$ $=16$

$x^2+4x-12$ $=0$

$x^2+6x-2x-12=0$

$x(x+6)-2(x+6)=0$

$(x+6)(x-2)=0$

$x=-6,2$

x cannot be negative

when $x=2$

$y^2=8\left(2\right)$

$y^2=16$

$y=4,-4$

so , The coordinates are $(2,4)and(2,-4)$