Answer to Question #99678 in Calculus for kawthar

Question

Find the tangents and normals to the curve
xy+2x-y=0
that are parallel to the line
2x+y=0
Draw the graphs showing the tangents and normal to the curve.

Answer 1

Let's rewrite the equation in form

$2x=y(1-x)$

Hence

$y=\frac{2x}{1-x}$

The equation of the tangent

$y_t=y'(x_0)(x-x_0)+y(x_0)$

$y'=\frac{2(1-x)+2x}{(1-x)^2}=\frac{2}{(1-x)^2}$

Since the tangent is parallel to the line

2x+y=0 

$y'(x_0)=-2\iff \frac{2}{(1-x)^2}=-2$

The last equation has no real roots hence there are no tangents to this curve which are parallel to this line.

The equation of the normal

$y_n=-\frac{1}{y'(x_0)}(x-x_0)+y(x_0)$

Since the normal is parallel to the line

2x+y=0 

$-\frac{1}{y'(x_0)}=-2\iff -\frac{(1-x^2)}{2}=-2$

$(1-x)^2=4$

$x_1=3;\quad x_2=-1$

Two normals to the curve have the equation

$y_{n1}=-2(x-3)-3=-2x+3$

$y_{n2}=-2(x+1)-1=-2x-3$