Question

Show that the following set of curves intersect orthogonally : 

y = x³ and 6y = 7 – x²

Answer 1

Given:

Curves y = x³ ...(1)

& 6y = 7 – x² ...(2)

Solving (1) & (2),we get

⇒ 6y = 7 – x²

⇒ 6(x³) = 7 – x²

⇒ 6x³ + x² – 7 = 0

Since f(x) = 6x³ + x² – 7,

we have to find f(x) = 0,so that x is a factor of f(x).

when x = 1

f(1) = 6(1)³ + (1)² – 7

f(1) = 6 + 1 – 7

f(1) = 0

Hence, x = 1 is a factor of f(x).

Substituting x = 1 in y = x³ ,we get

y = 1³

y = 1

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

First curve y = x³

Differentiating above w.r.t x,

⇒ m₁ = \(\frac{dy}{dx}\) = 3x² ...(3)

Second curve 6y = 7 – x²

Differentiating above w.r.t x,

⇒ 6 \(\frac{dy}{dx}\) = 0 – 2x

⇒ m₂ = \(\frac{-2x}{6}\)

⇒ m₂ = \(\frac{-x}{3}\)...(4)

At (1,1),we have,

m₁ = 3x²

⇒ 3×(1)²

m₁ = 3

At (1,1),we have,

⇒ m₂ = \(\frac{-x}{3}\)

⇒ \(\frac{-1}{3}\)

⇒ m₂ = \(\frac{-1}{3}\)

When m₁ = 3 & m₂ = \(\frac{-1}{3}\)

⇒ 3 × \(\frac{-1}{3}\) = -1

∴ Two curves y = x³ & 6y = 7 – x² intersect orthogonally.