Question

Find the point of intersections of the tangets drawn to the curve $x^{2}y=1-y$ at the points where it is intersected by the curve $xy=1-y$

• A. $(1,0)$
• B. $(0,1)$
• C. $(\frac{2}{3},\frac{1}{3})$
• D. $(\frac{1}{3},\frac{2}{3})$

Answer 1

Answer: (B), (D)

$(0,1)$
$(\frac{1}{3},\frac{2}{3})$

Given equation of curves
$x^{2}y=1-y$ .....(1)
$xy=1-y$ ....(2)
Let $P(x_1,y_1)$ be the point of intersection of the curves.
$\Rightarrow x_1^2{y}_1=x_1{y}_1$
$\Rightarrow x_1{y}_1(x_1-1)=0$
$\Rightarrow x_1=0,x_1=1$
$\Rightarrow y_1=1,\frac{1}{2}$
So, the point of intersection of the curves are $(0,1)$ and $(1,\frac{1}{2})$
Slope of tangent to curve (1),
$2xy+x^2{y}^{\prime }=-y^{\prime }$
$\Rightarrow y^{\prime }=\frac{-2xy}{x^2+1}$
Slope of tangent to curve (1) at $(0,1)$ is $0.$
Slope of tangent to curve (1) at $(1,\frac{1}{2})$ is $-\frac{1}{2}$
Equation of tangent to curve (1) at $(0,1)$ is
$y-1=0$ .....(3)
Equation of tangent to curve (1) at $(1,\frac{1}{2})$ is
$y-\frac{1}{2}=-\frac{1}{2}(x-1)$
$\Rightarrow 2y+x-2=0$ ....(4)
Slope of tangent to curve (2),
$x{y}^{\prime }+y=-y^{\prime }$
$y^{\prime }=-\frac{y}{1+x}$
Slope of tangent to curve (2) at $(0,1)$ is $-1$.
Slope of tangent to curve (2) at $(1,\frac{1}{2})$ is $-\frac{1}{4}$
Equation of tangent to curve (2) at $(0,1)$ is
$y-1=-1(x-0)$
$\Rightarrow y+x-1=0$ .....(5)
Equation of tangent to curve (2) at $(1,\frac{1}{2})$ is
$y-\frac{1}{2}=-\frac{1}{4}(x-1)$
$\Rightarrow x+4y-3=0$ ....(6)
Solving (3) and (5),we get
$x=0,y=1$
Solving (4) and (6), we get
$x=\frac{1}{3},y=\frac{2}{3}$
So, the required points are $(0,1)$ and $(\frac{1}{3},\frac{2}{3})$