Respuesta :
Answer:
[tex]m=\frac{9}{13}[/tex] and [tex]b=\frac{40}{13}[/tex]
Step-by-step explanation:
The equation of curve is
[tex]2(x^2+y^2)^2=25(x^2-y^2)[/tex]
We need to find the equation of the tangent line to the curve at the point (-3, 1).
Differentiate with respect to x.
[tex]2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})[/tex]
[tex]4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})[/tex]
The point of tangency is (-3,1). It means the slope of tangent is [tex]\frac{dy}{dx}_{(-3,1)}[/tex].
Substitute x=-3 and y=1 in the above equation.
[tex]4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})[/tex]
[tex]40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})[/tex]
[tex]-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}[/tex]
[tex]80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240[/tex]
[tex]130\frac{dy}{dx}=90[/tex]
Divide both sides by 130.
[tex]\frac{dy}{dx}=\frac{9}{13}[/tex]
If a line passes through a points [tex](x_1,y_1)[/tex] with slope m, then the point slope form of the line is
[tex]y-y_1=m(x-x_1)[/tex]
The slope of tangent line is [tex]\frac{9}{13}[/tex] and it passes through the point (-3,1). So, the equation of tangent is
[tex]y-1=\frac{9}{13}(x-(-3))[/tex]
[tex]y-1=\frac{9}{13}(x)+\frac{27}{13}[/tex]
Add 1 on both sides.
[tex]y=\frac{9}{13}(x)+\frac{27}{13}+1[/tex]
[tex]y=\frac{9}{13}(x)+\frac{40}{13}[/tex]
Therefore, [tex]m=\frac{9}{13}[/tex] and [tex]b=\frac{40}{13}[/tex].
