Question

Find the equation of the line with slope equal to 1 and normal to the
curve x
2 + 3xy + y
2 = 5 at P0(x0, y0) when x0 = 1.

Answer 1

Solution;

Given the equation of the curve is;

$x^2+3xy+y^2=5$

At $P_0(x_0,y_0)$ When $x_0=1$

Differentiate as follows;

$2x+3y+3x\frac{dy}{dx}+2y\frac{dy}{dx}=0$

Group;

$(3x+2y)\frac{dy}{dx}=-(2x+3y)$

$\frac{dy}{dx}=\frac{-(2x+3y)}{3x+2y}$

Since $x_0=1$ ,by direct substitute into the equation,find values of y.

$x^2+3xy+y^2=5$

$(1^2)+3(1)y+y^2=5$

$y^2+3y-4=0$

$y^2-y+4y-4=0$

$y(y-1)+4(y-1)=0$

Hence ;

$y=1 or -4$

Use y=1 because at y=-4,the gradient of the equation is not equal to 1.

So the point $P_0(x_0,y_0)=P_0(1,1)$

$\frac{dy}{dx}(1,1)=\frac{-(2+3)}{3+2}=-1$

The equation of the normal to curve is ;

$(x-1)=1(y-1)$

$x-y=0$

This is the required solution.