Question

Tangent is drawn at any point $P$ of a curve which passes through $(1,1)$ cutting $x$-axis and $y$-axis at $A$ and $B$, respectively. If $AP:BP=3:1$, then

• A. differential equation of the curve is $3x\frac{dy}{dx}+y=0$.
• B. differential equation of the curve is $3x\frac{dy}{dx}-y=0$.
• C. curve is passing through $\left(\frac{1}{8},2\right)$
• D. normal at $(1,1)$ is $x+3y=4$.

Answer 1

Answer: (A), (C)

The equation of tangent to any curve at point $P(x,y)$ is
$Y-y=\frac{dy}{dx}(X-x)$
which cuts the $x$-axis at $Y=0$ :
$X=x-y\frac{dy}{dx}$
that is, at
A $\left(x-y\frac{dy}{dx},0\right)$
and it also cuts the $y$-axis at $X=0$ :
$Y=y-x\frac{dy}{dx}$
That is, at
B $\left(0,y-x\frac{dy}{dx}\right)$

$x=\frac{\left[\left(x-y\frac{dy}{dx}\right)\times 1\right]+(3\times 0)}{1+3}$
$4x=x-y\frac{dy}{dx}\Rightarrow y\frac{dx}{dy}=-3x$
$y=\frac{(1\times 0)+\left[3\left(y-x\frac{dy}{dx}\right)\right]}{4}$
$\Rightarrow 4y=3y-3x\frac{dy}{dx}$
$\Rightarrow 3x\frac{dy}{dx}=-y$
The equation of the curve is
$3x\frac{dy}{dx}+y=0$
$3\int \frac{dy}{y}+\int \frac{dy}{x}=\ln c$
$3\ln y+\ln x=\ln c$
$x{y}^3=c$
which is passing through the point $(1,1).$
Therefore, $c=1$ and hence
$x{y}^3=1$
which is the equation of curve that passes through $(1/8,2)$.