JEE Advanced · 2006 · 09 Apr 2006 · Differential Equations

A tangent drawn to the curve

y = f (x)

P (x, y)

cuts the x-axis and y-axis at A and B respectively such that BP : AP

= 3 : 1

, given that

f (1) = 1

, then

Answer: (C), (D)

Solution

The tangent at

P (x, y)

has equation:

Y - y = \frac{d y}{d x} (X - x)

This meets the x-axis at

A (x - \frac{y}{d y / d x}, 0)

and the y-axis at

B (0, y - x \frac{d y}{d x})

.

Given

\frac{B P}{A P} = \frac{3}{1}

, using the section formula and the geometry of the tangent line, we get:

\frac{d x}{x} = - \frac{d y}{3 y}

Integrating:

ln x = - \frac{1}{3} ln y + C

x^{3} = \frac{k}{y}

Using

f (1) = 1

1 = \frac{k}{1}

, so

k = 1

.

Therefore:

y = \frac{1}{x ^{3}}

This can be written as the differential equation:

x \frac{d y}{d x} + 3 y = 0

Checking

(2, 1/8)

y = \frac{1}{2 ^{3}} = \frac{1}{8}

✓

So options (C) and (D) are correct.

JEE Advanced · 2006 · 09 Apr 2006 · One or more correct options