Answer to Question #294487 in Calculus for Ianidy

Question

show that the equation of the tangent to x²+xy+y=0 at the point(x₁,y₁) is

(2x₁+y₁)x+(x₁+1) y+y₁ =0

Answer 1

$x^2+xy+y=0$

Differentiate both sides with respect to $x$

$\dfrac{d}{dx}(x^2+xy+y)=\dfrac{d}{dx}(0)$

Use the Chain Rule

$2x+y+x\dfrac{dy}{dx}+\dfrac{dy}{dx}=0$

Solve for $\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=-\dfrac{2x+y}{x+1}$

At the point $(x_1,y_1)$

$\dfrac{dy}{dx}|_{(x_1,y_1)}=-\dfrac{2x_1+y_1}{x_1+1}$

The equation of the tangent to $x^2+xy+y=0$ at the point$(x_1,y_1)$ is

$y-y_1=-\dfrac{2x_1+y_1}{x_1+1}(x-x_1)$

$(x_1+1)y+(2x_1+y_1)x-x_1y_1-y_1 -2x_1^2-x_1y_1=0$

$(x_1+1)y+(2x_1+y_1)x-2(x_1^2+x_1y_1+y_1)+y_1=0$

Then the equation of the tangent to $x^2+xy+y=0$ at the point$(x_1,y_1)$ will be

$(x_1+1)y+(2x_1+y_1)x+y_1=0$