Question

The sum of squares of intercepts on the axes cut off by the tangents to the curve x^(2/3) + y^(2/3) = a^(2/3) (a > 0) at [(a/8), (a/8)] is 2. Thus a has the value. 

(a) 1

(b) 2

(c) 4

(d) 8   

Answer 1

The correct option (c) 4  

Explanation:

 x^(2/3) + y^(2/3) = a^(2/3)

∴ [(2/3)x^–(1/3)] + [(2/3)y^–(1/3)](dy/dx) = 0

∴ x^–(1/3) + y^–(1/3)(dy/dx) = 0

∴ (dy/dx) = [(– x^–(1/3))/(y^–(1/3))] = [(– y^–(1/3))/(x^(1/3))] 

∴ (dy/dx) = [(– y)/x]^1/3

∴ (dy/dx) at [(a/8), (a/8)] is (dy/dx) = – 1

∴ equation of tangent at [(a/8), (a/8)] is

y – (a/8) = – [x – (a/8)] i.e.

x + y – (a/4) = 0

∴ sum of intercepts = (a/4) + (a/4) = (a/2)

i.e. 2 = (a/2)

∴ a = 4