Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A the differential equation of the curve is $3x\frac{dy}{dx}+y=0$
C the curve passes through $(\frac{1}{8},2)$

Since $BP:AP=3:1$
Equation of the tangent is $Y-y=f^{\prime }\left(x\right)(X-x).$
Intercept on $X$-axis $(x-\frac{y}{f^{\prime }\left(x\right))},0)$, Y-axis$(0,y-x{f}^{\prime }(x\left)\right)$.
Now $x=\frac{(x-\frac{y}{f^{\prime }\left(x\right)})+1\times 0}{3+1}$ {since, it divides internally $3:1$}
$\Rightarrow \frac{dy}{dx}=-\frac{y}{3x}\Rightarrow \frac{dy}{y}=-\frac{dx}{3x}\Rightarrow 3x\frac{dy}{dx}+y=0$
$\Rightarrow \log y=-\frac{1}{3}\log x+\log C\Rightarrow x{y}^3=C$
But curve passes through $(1,1)$ $\Rightarrow 1=C$
$\therefore x{y}^3=1$
$\therefore$curve passes through $(\frac{1}{8},2)$.