Question

A circle circumscribing the triangle formed by three co−normal points passes through the vertex of the parabola and its equation is, 2(x² + y²) − 2(h + 2a)x − ky = 0.

Answer 1

Equation of the normal at P 

y + tx = 2at + at³ 

passes through (h, k)

Let the circle through PQR is 

x² + y² + 2gx + 2fy + c = 0

solving circle x = at² , y = 2at

⇒ circle passes through the origin 

hence the equation of the circle 

x² + y² + 2gx + 2fy = 0 

now equation (2) becomes 

(1) and (3) must have the same root 

2(2a + g) = 2a – h 

2g = – (h + 2a)

and 4f = – k

=> 2f = - k/2

Hence the equation of the circle is

x² + y² – (h + 2a)x – k/2 y = 0

=> 2(x² + y²) − 2(h + 2a)x − ky = 0 ]