Show that the following set of curves intersect orthogonally :
y = x3 and 6y = 7 – x2
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 = 3x2 ...(3)
Second curve 6y = 7 – x2
Differentiating above w.r.t x,
⇒ 6 = 0 – 2x
⇒ m2
⇒ m2 ...(4)
At (1,1),we have,
m1 = 3x2
⇒ 3×(1)2
m1 = 3
At (1,1),we have,
⇒ m2
⇒
⇒ m2
When m1 = 3 & m2 =
⇒ 3×1
∴ Two curves y = x3 & 6y = 7 – x2 intersect orthogonally.