Question

The curve `(x/a)^n + (y/b)^n` = 2, touches the line `x/a + y/b` = 2 at the point (a, b) for n is equal to ______.

Options

• 1

• 2

• 3

• all non zero values of n

Answer 1

The curve `(x/a)^n + (y/b)^n` = 2, touches the line `x/a + y/b` = 2 at the point (a, b) for n is equal to all non zero values of n.

Explanation:

`(x/a)^n + (y/b)^n` = 2,

Differentiating above equation w.r.t.x, we get

`(nx^(n-1))/a^n + (ny^(n - 1))/b^n xx (dy)/(dx)` = 0

`(ny^(n - 1))/b^n xx (dy)/(dx) = -(nx^(n - 1))/a^n`

⇒ `(dy/dx) = -(nx^(n - 1))/a^n xx b^n/(ny^(n - 1))`

`(dy/dx) = - b^n/a^n xx (x/y)^(n - 1)`

`(dy/dx) = b^n/a^n xx (a/b)^(n - 1)`

= `-b^n/a^n xx a^n/b^n xx b/a`

`(dy)/(dx)|_((a","  b)) = - b/a`

Compare slope of given tangent slope = `-b/a`

Hence, it is valid for all values of n ≠ 0.