Tangents at the end of focal chord are perpendicular to each other
Slope of tangent at Q= Slope of normal at P
Slope of normal to y2=4ax at P is −t
Given, x=y2−8y+2
(y−4)2=(x+14)
Any point on the parabola will be (−14+t24,4+t2)
On comparing it with (−5,1), we get t=−6
∴ Slope of the normal at P will be =6
Hence, Slope of tangent at Q≡6