Question

If the tangent and normal is drawn at the point(s) of intersection of the curves $x^2{y}^2=1-x$ and $xy=1-x,$ then which of the following(s) is /are correct for first curve

• A. Tangent to first curve is parallel to $x-$axis
• B. Tangent to first curve is parallel to $y-$axis
• C. Normal to first curve is parallel to $x-$axis
• D. Normal to first curve is parallel to $y-$axis

Answer 1

Answer: (B), (C)

Normal to first curve is parallel to $x-$axis

Given curves : $x^2{y}^2=1-x$ and $xy=1-x$
For point of intersection
$x^2{y}^2=xy$
$\Rightarrow xy(xy-1)=0$
$\Rightarrow x=0$ or $y=0$ or $xy=1$
But $x=0$ does not satisfy first curve (rejected)
For $y=0\Rightarrow x=1$
and
For $xy=1$
$\Rightarrow 1=1-x,x=0$ (rejected)
So, only one point of intersection as $(1,0)$

Now, $x^2{y}^2=1-x$
differentiating both sides w.r.t. $x$
$\Rightarrow 2x{y}^2+2x^{2}y{y}^{\prime }=-1$
$\Rightarrow y^{\prime }=-\frac{1+2x{y}^2}{2x^{2}y}=\frac{dy}{dx}$
Slope of tangent at $(1,0)$
$\Rightarrow \frac{dy}{dx}{|}_{(1,0)}=-\infty$
$\Rightarrow$ slope of normal $=0$
$\therefore$ Normal is parallel to $x-$axis and tangent is parallel to $y-$axis