The tangent and normal at the point P ( 4 , 4 ) to the parabola, y 2 = 4 x intersect the x -axis at the points Q and R respectively. Then the cirucm centre of the Δ PQR is

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Eq. of tangent 2 y = x + 4 ∴ Q ≡ ( − 4 , 0 ) Eq. of normal is y − 4 = − 2 ( x − 4 ) ⇒ y + 2 x = 12 Clearly QR is diameter of the required circle. ​ ⇒ ( x + 4 ) ( x − 6 ) + y 2 = 0 ​ ⇒ x 2 + y 2 − 2 x − 24 = 0 ​ centre ( 1 , 0 )

Mathematics - Parabola Question with Solution | TestHub

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