Question

for the parabola y²=4x, find the coordinate of the point whose focal distance is 17.​

Answer 1

The Coordinate of point is (16, 8)

1) Given Equation of parabola is $y^2 =4x$.

2) General Equation of a parabola is $y^2 = 4ax$.

Comparing this two equations gives a = 1.

3) So, the position of focus of the parabola is (1,0).

4) Let the co-ordinates of the point be( c, d ).

5) The focal distance of point is given to be 17. i.e distance between (c,d) and (1,0) is 17 units.

6) Using the distance formula $\sqrt{(c-1)^2 + (d-0)^2 } =17$.

7) Simplifying,  $\sqrt{(c-1)^2 + d^2} = 17$.

8) Since, the point(c,d) lies on the parabola so, $d^{2} = 4 c$.

9) Substituting value of d² in the distance equation gives $\sqrt{(c-1)^2 +4c} =17$

10)  This can be simplified to $\sqrt{(c+1)^2} =17$.

11)  c = -18, 16.

12) Since $d^2 = 4c$, so we can't have negative value for c.

13) Hence c = 16, and d = 8.