The equation of parabola is, y2=8x Differentiate the above equation with respect to x.
dy
dx
=
4
y
At point P(2,4), the slope of tangent is, m1=
4
4
=1 The equation of 1st tangent is, y−4=1(x−2) y=x+2 …… (1) At point Q(18,−12), the slope of tangent is, m2=4−12 =−
1
3
The equation of 2nd tangent is, y+12=−
1
3
(x−18) 3y+18=−x …… (2) From equation (1) and equation (2), y=−4 x=−6 So, the tangents intersect at (-6,-4) The required equation of tangent passes through (-6,-4) andhaving a slope