Question

The differential equation y$dx+y^{2}dy=xdy$ and $y\left(1\right)=1$ represents a parabola whose

• A. Vertex is (0, 0)
• B. Axis is x = 1
• C. Length of latus rectum is 4
• D. Tangent at the vertex is x= 1

Answer 1

The correct option is D Tangent at the vertex is x= 1

Given eqn. in $\frac{ydx-xdy}{y^2}+dy=0\Rightarrow d\left(\frac{x}{y}\right)+dy=0$
$\Rightarrow \frac{x}{y}+y=c$, but $y\left(1\right)=1\Rightarrow C=2$
$\therefore \frac{x}{y}+y=2\Rightarrow x+y^2=2y\Rightarrow (y-1{)}^2=-(x-1)$