Question 1040185
Find the derivative.
{{{2xdx+2ydy=0}}}
{{{2ydy=-2xdx}}}
{{{dy/dx=-x/y}}}
So when {{{y=-x}}}, the derivative (which is the slope of the tangent line) is equal to 1.
To find the points, plug this into the original equation.
{{{x^2+(-x)^2=9}}}
{{{2x^2=9}}}
{{{x^2=9/2}}}
{{{x=0 +- 3/sqrt(2)}}}
{{{x=0 +- (3/2)sqrt(2)}}}
Then using the point-slope form,
{{{y-(3/2)sqrt(2)=1(x+(3/2)sqrt(2))}}}
{{{y=x+3sqrt(2)}}}
and
{{{y+(3/2)sqrt(2)=1(x-(3/2)sqrt(2))}}}
{{{y=x-3sqrt(2)}}}
.
.
.
*[illustration xc4.JPG].