Question

A curve $y=f\left(x\right)$ passes through $(1,1)$ and tangent at $P(x,y)$ cuts the $x-$axis and $y-$axis at $A$ and $B$ respectively such that $AP:PB=1:3,$ then which of the following is/are true:

• A. Differential equation of curve is $x{y}^{\prime }-3y=0$
• B. Differential equation of curve is $x{y}^{\prime }+3y=0$
• C. Curve passes through $(2,\frac{1}{8})$
• D. Curve passes through $(2,\frac{1}{4})$

Answer 1

Answer: (B), (C)

Curve passes through $(2,\frac{1}{8})$

Tangent to the curve $y=f\left(x\right)$ at $P(x,y)$ is
$Y-y=\frac{dy}{dx}(X-x)$
$\therefore A(\frac{x\frac{dy}{dx}-y}{\frac{dy}{dx}},0);B(0,-x\frac{dy}{dx}+y)$
$\because AP:PB=1:3$
Using section formula, we have:
$x=\frac{\frac{3(x\frac{dy}{dx}-y)}{\frac{dy}{dx}}+1\times 0}{4}$
$\Rightarrow x\frac{dy}{dx}+3y=0$
( Note that the same we will get with $y-$co-ordinate)
Which can also be written as $x{y}^{\prime }+3y=0$
$\Rightarrow \int \frac{dy}{y}=\int -3\frac{dx}{x}$
$\Rightarrow \ln |y|=-3\ln |x|+\ln |c|$
$\Rightarrow y=\frac{c}{x^3}$
As curve passes through $(1,1),c=1$
$\therefore$ curve is $x^{3}y=1$ which also passes through $(2,\frac{1}{8})$