Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.
Given:
Curves xy = 4 ...(1)
& x2 + y2 = 8 ...(2)
Solving (1) & (2),we get
⇒ xy = 4
⇒ x
Substituting x in x2 + y2 = 8,we get,
⇒ ()2 + y2 = 8
⇒ + y2 = 8
⇒ 16 + y4 = 8y2
⇒ y4 – 8y2 + 16 = 0
We will use factorization method to solve the above equation
⇒ y4 – 4y2 – 4y2 + 16 = 0
⇒ y2(y2 – 4) – 4(y2 – 4) = 0
⇒ (y2 – 4)(y2 – 4) = 0
⇒ y2 – 4 = 0
⇒ y2 = 4
⇒ y = ±2
Substituting y = ±2 in x,we get,
⇒ x
⇒ x = ±2
∴ The point of intersection of two curves (2,2) &
( – 2, – 2)
First curve xy = 4
⇒ 1×y + x. = 0
⇒ x. = – y
⇒ m1 ...(3)
Second curve is x2 + y2 = 8
Differentiating above w.r.t x,
⇒ 2x + 2y. = 0
⇒ y. = – x
⇒ m2 ...(4)
At (2,2),we have,
m1
⇒
m1 = – 1
At (2,2),we have,
⇒ m2
⇒
⇒ m2 = – 1
Clearly, m1 = m2 = – 1 at (2,2)
So, given curve touch each other at (2,2)