Question

find focal distance of the point P(4 , -8) on the parabola y² = 16x​

Answer 1

General equation of parabola is :-

$\boxed{\rm \large \: {y}^{2} = 4 ax}....(1)$

But in the question , equation of parabola is given as :-

y² = 16x..... (2)

Now by comparing (1) and (2) :-

$\rm \rightarrow \: 4ax = 16x \\ \\ \\ \rm \rightarrow \: 4a = 16\\ \\ \\ \rm \rightarrow \: a = \dfrac{16}{4} \\ \\ \\ \rm \rightarrow \: a = 4$

Coordinates of vertex of given parabola :- (0,0)

coordinates of Focus :- (4,0)

Now , we have to find out distance between point (4,0) and P(4 , -8).

$\implies \sqrt{ {(4 - 4)}^{2} +{(0 + 8)}^{2} } \\ \\ \\ \implies \sqrt{ 0+{( 8)}^{2} }\\ \\ \\ \implies \sqrt{ {( 8)}^{2} } \\ \\ \\ \implies \sqrt{ 64 }\\ \\ \\ \implies 8$

Focal distance of the point P(4 , -8) on the parabola y² = 16x = 8 units

Answer 2

Hope the above attachment helps u this answer was not copied