Question

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

(a) (0, −1),
(b) (−1/2, 1/2)

Answer 1

a.
`y+1=0`

b.
`2x+4y=1`

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=(-3+2)/(2)=-(1)/(2)`

`m=y'=-1/2`

Again using point-slope formula

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

`(2y-1)/(2)=-(1)/(4)(2x+1)`

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

`4y-2=-2x-1`

`2x+4y=2-1`

`2x+4y=1`