Find the value of p for which the curves x²=9p(9-y) and x²=p(y+1) cut each other at right angles.

The given curves are
x²=9p(9-y) (1)
x²=p(y+1). (2)
Solving (1) and (2) simultaneously, we get
9p(9-y)=p(y+1)
⇒81p - 9py = py + p
⇒81p - p = 10py
⇒80p = 10py
⇒10y = 80
⇒y = 8
x²=9p(9-8) ⇒x²=9p ⇒x=±3√p
Therefore, given curves intersect at points(±3√p,8) i.e., at points P(3√p,8) and Q(-3√p,8)
Differentiating (1) and (2) w.r.t x, we get
2x=-9p dy/dx ⇒ dy/dx = -2x/9p. (3)
and 2x=p dy/dx ⇒ dy/dx = 2x/p. (4)
Slope of tangent to the curve (1) at point P is m₁ = -2(3√p)/9p = -2/(3√p)
Slope of tangent to the curve (2) at point P is m₂ = 2(3√p)/p = 6/√p
Now given curves (1) and (2) cut at right angles,
Therefore, m₁ × m₂ = -1
⇒[-2/(3√p)] × [6/√p] = -1
⇒-12/(3p) = -1
⇒-4/p = -1
⇒4 = p
⇒p = 4
Similarly, when the curves intersect at right angles at point Q, then p=4.