The given equation of the parabola is y2=4x Equation of tangent at (1,2) is ⇒y⋅2=2(x+1) ⇒y=x+1 Slope of tangent =1 ∴ slope of the normal =−1 Equation of the normal passing through (1,2) is ⇒y−2=−1(x−1) ⇒y−2=−x+1⇒x+y=3 This equation again meet at Q of the parabola, then y2=4(3−y) ⇒y2=12−4y⇒y2+4y−12=0 ⇒(y−2)(y+6)=0 ⇒y=2,−6 If y=−6, then x=3−(−6)=9 ∴ The coordinates of point Q are (9,−6).