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 $A P: B P=3: 1$, then

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

Answer 1

Answer: (A), (C)

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

$ x =\frac{\left[\left(x-y \frac{d y}{d x}\right) \times 1\right]+(3 \times 0)}{1+3} $
$4 x = x - y \frac{d y}{d x} \Rightarrow y \frac{d x}{d y}=-3 x $
$y =\frac{(1 \times 0)+\left[3\left(y-x \frac{d y}{d x}\right)\right]}{4}$
$ \Rightarrow 4 y =3 y -3 x \frac{d y}{d x} $
$\Rightarrow 3 x \frac{d y}{d x}=- y$
The equation of the curve is
$3 x \frac{d y}{d x}+y=0$
$3 \int \frac{d y}{y}+\int \frac{d y}{x}=\ln c$
$3 \ln y+\ln x=\ln c$
$x y^{3}=c$
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 $(1 / 8,2)$.