ISI-2004-30

Question

If the tangent at the point $P$ with co-ordinates $(h, k)$ on the curve $y^2 = 2x^3$ is perpendicular to the straight line $4x=3y$, then

• A. $(h, k) = (0,0)$
• B. $(h, k) = (1/8, -1/16)$
• C. $(h, k) = (0,0)$ or $(h, k) = (1/8, -1/16)$
• D. no such point $(h, k)$ exists.

Comments

y² = 2x³

Differentiating both sides wrt 'x'

2y(dy/dx) = 6x²

^So, dy/dx = 3x²/y

So, this is the slope of tangent on the given curve.

Now ,tangent at point P has co-ordinates (h,k) .

So, it will satisfy dy/dx = 3x²/y

So , let ,m₁  = dy/dx = 3h²/k

Since we know that 2 lines are perpendicular when product of the slopes of 2 lines is -1

i.e. m₁*m₂  = -1

Here , m₂ = 4/3

So , (3h²/k)* (4/3) = -1

k = -4h²

 Now ,check options...

C is satisfying k = -4h².

So, C should be the answer