Question

Show that the following curves intersect orthogonally at the indicated points : 

x² = 4y and 4y + x² = 8 at (2, 1)

Answer 1

Given:

Curves x² = 4y ...(1)

& 4y + x² = 8 ...(2)

The point of intersection of two curves (2,1)

Solving (1) & (2),we get,

First curve is x² = 4y

Differentiating above w.r.t x,

⇒ 2x= 4.\(\frac{dy}{dx}\)

⇒ \(\frac{dy}{dx}\) = \(\frac{2x}{4}\)

⇒ m₁ = \(\frac{x}{2}\)...(3)

Second curve is 4y + x² = 8

⇒ 4.\(\frac{dy}{dx}\) + 2x = 0

 ⇒ \(\frac{dy}{dx}\) = \(\frac{-2x}{4}\)

⇒ m₂ = \(\frac{-x}{2}\)...(4)

Substituting (2,1) for m₁ & m₂,we get,

m₁ = \(\frac{x}{2}\)

⇒ \(\frac{2}{2}\)

m₁ = 1 ...(5)

m₂ = \(\frac{-x}{2}\)

⇒ \(\frac{-2}{2}\)

m₂ = – 1 ...(6)

when m₁ = 1 & m₂ = – 1

⇒ 1× – 1 = – 1

∴ Two curves x² = 4y & 4y + x² = 8 intersect orthogonally.