Answer to Question #183279 in Calculus for Zannat

Question

Find all The points on the graph of (x^2+y^2+y)^2=x^2+y^2 that have horizontal tangent line

Answer 1

Given, the function (x²+y²+y)²=x²+y². ...................................(1)

First, differentiate with respect to x.

2(x²+y²+y)(2x+2yy'+y')=2x+2yy'

Second, putting y'=0.

4x(x²+y²+y)=2x

2(x²+y²+y)=1

x²+y²+y=1/2 .............................................................(2)

This is the horizontal tangent line.

Third, substitute the above horizontal tangent line in the given function.

i.e., putting equation (2) in equation (1),

(x²+y²+y)²=x²+y²

(1/2)²= -y+1/2 { from (2), x²+y²= -y+1/2 }

1/4= -y+1/2

y=1/4

Forth, finding the x-coordinate.By putting the value of y in the horizontal tangent line.

x=$\pm\frac{\sqrt{3}}{4}$

Thus, the point are $(\frac{-\sqrt3}{4},\frac{1}{4})$ and $(\frac{\sqrt3}{4},\frac{1}{4})$ on the given graph that have horizontal tangent line.