Question

A curve y = f(x) passes through (1, 1) and the tangent at P(x, y) cuts the x-axis and y-axis at A and B, respectively such that BP : AP = 3 : 1, then 

(a) the equation of the curve is xy' – 3y = 0. 

(b) the normal at (1, 1) is x + 3y = 4. 

(c) the curve passes through (2, 1/8). 

(d) the equation of the curve is xy' + 3y = 0.

Answer 1

Answer is (b) the normal at (1, 1) is x + 3y = 4.

Explanation:

The equation of any tangent at P(x, y) is

Y – y = (dy/dx)(X – x)

It meets the x-axis at A(x – y(dx/dy), 0) and B(0, y – x(dy/dx))

Since P(x, y) divides the line AB in the ratio 3 : 1, we get

log |y| = log c – 3log |x|

x³y = c 

As f(1) = 1, we get c = 1

Thus, the equation of the curve is x³y = 1

3y – 3 = x – 1 

x – 3y + 2 = 0