Question

A circle having its center at (2, 3) is cut orthogonally by the parabola y2=4x. The possible intersection point(s) of these curves, can be

Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

Completing the question :

Find the point of intersection where parabola cuts circle orthoginally. (only)

Solution :

Final Answer : (4,4) ,(9,6)

Steps and Understanding :

1) Point where it cuts the graph orthogonally is where tangent at parabola is passing through centre or normal to circle.

Let that point be in parametric form (t^2 ,2t) .

2) Tangent at that point :

3) Centre : (6,5)

Now, slope of normal is :

For orthogonal,

4) Point where orthogonality achieves are :

(t^2 , 2t) = (4,4) & (9,6)