Question

If the parabola y^2=4x meets a circle with centre at (6,5)orthogonally

Answer 1

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 :
${y}^{2} = 4x \\ = > 2y \: dy = 4 \: dx \\ = > \frac{dy}{dx} = \frac{2}{y} = \frac{2}{2t} = \frac{1}{t}$

3) Centre : (6,5)
Now, slope of normal is :
$\frac{ 2t - 5 }{ {t}^{2} - 6 }$
For orthogonal,
$\frac{2t - 5}{ {t}^{2} - 6} = \frac{1}{t} \\ = > {t}^{2} - 5t + 6 = 0 \\ = > (t - 3)(t - 2) = 0 \\ = > t = 2 \: or \: t \: = 3$

4) Point where orthogonality achieves are :
(t^2 , 2t) = (4,4) & (9,6)