Answer:
a.[tex]y+1=0[/tex]
b.[tex]2x+4y=1[/tex]
Step-by-step explanation:
We are given that
[tex](x^2+y^2)^2=3x^2y-y^3[/tex]
a.(0,-1)
Differentiate w.r.t x
[tex]2(x^2+y^2)(2x+2yy')=6xy+3x^2y'-3y^2y'[/tex].....(1)
Substitute x=0 and y=-1 in equation (1)
[tex]2(0+1)(-2y')=-3y'[/tex]
[tex]-4y'+3y'=0[/tex]
[tex]-y'=0[/tex]
[tex]y'=0[/tex]
[tex]m=y'=0[/tex]
Point-slope form:
[tex]y-y_0=m(x-x_0)[/tex]
Using the formula
[tex]y+1=0[/tex]
This is required equation of tangent line to the given curve at point (0,-1).
b.(-1/2,1/2)
Substitute the value in equation (1)
[tex]2(1/4+1/4)(-1+y')=6(-1/2)(1/2)+3(1/4)y'-3(1/4)y'[/tex]
[tex]2(2/4)(-1+y')=-3/2+3/4y'-3/4y'[/tex]
[tex]-1+y'=-3/2[/tex]
[tex]y'=-3/2+1=\frac{-3+2}{2}=-\frac{1}{2}[/tex]
[tex]m=y'=-1/2[/tex]
Again using point-slope formula
[tex]y-1/2=-1/2(x+1/2)[/tex]
[tex]\frac{2y-1}{2}=-\frac{1}{4}(2x+1)[/tex]
[tex]2y-1=-\frac{1}{2}(2x+1)[/tex]
[tex]4y-2=-2x-1[/tex]
[tex]2x+4y=2-1[/tex]
[tex]2x+4y=1[/tex]
