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Rina8888 [55]
3 years ago
13

Use implicit differentiation to find an equation of the tangent line to the curve at the given points. (x2 + y2)2 = 3x2y − y3;

Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

a.y+1=0

b.2x+4y=1

Step-by-step explanation:

We are given that

(x^2+y^2)^2=3x^2y-y^3

a.(0,-1)

Differentiate w.r.t x

2(x^2+y^2)(2x+2yy')=6xy+3x^2y'-3y^2y'.....(1)

Substitute x=0 and y=-1 in equation (1)

2(0+1)(-2y')=-3y'

-4y'+3y'=0

-y'=0

y'=0

m=y'=0

Point-slope form:

y-y_0=m(x-x_0)

Using the formula

y+1=0

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)

2(1/4+1/4)(-1+y')=6(-1/2)(1/2)+3(1/4)y'-3(1/4)y'

2(2/4)(-1+y')=-3/2+3/4y'-3/4y'

-1+y'=-3/2

y'=-3/2+1=\frac{-3+2}{2}=-\frac{1}{2}

m=y'=-1/2

Again using point-slope formula

y-1/2=-1/2(x+1/2)

\frac{2y-1}{2}=-\frac{1}{4}(2x+1)

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

4y-2=-2x-1

2x+4y=2-1

2x+4y=1

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