Question:
A curve $y=f(x)$ passes through $(1,1)$ and at $P(x,y),$ tangent cuts the $x$-axis and $y$-axis at $A$ and $B$ respectively such that $BP:AP=3:1$ then
A) $xy'-3y=0$
B) normal at $(1,1)$ is $x+3y=0$
C) curve passes through $(2,1/8)$
D) equation of curve is $xy'+3y=0$
My Attempt:
Let the tangent at $P$ is $Y-y=y'(X-x)$, where $y'$ is the slope of the curve at P.
Therefore $A$ is $(-y/y'+x,0)$ and $B$ is $(0,-xy'+y)$
Assuming internal division, I am getting $D$ option.
Assuming external division, I am getting $A$ option.
But as per answer key, option $A$ is not correct.
Why can't we take external division here?