Question

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

Answer 1

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

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

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

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

$4y-2=-2x-1$

$2x+4y=2-1$

$2x+4y=1$