Question 1107273
<br>
Use implicit differentiation to find an expression for dy/dx and show that there is a value of y for which the derivative is zero when x is 2.<br>
{{{4x^2+y^2=48+2xy}}}<br>
{{{8x+2y*(dy/dx) = 2x*(dy/dx)+2y}}}
{{{2y*(dy/dx) - 2x*(dy/dx) = 2y-8x}}}
{{{(2y-2x)*(dy/dx) = 2y-8x}}}
{{{dy/dx = (2y-8x)/(2y-2x) = (y-4x)/(y-x)}}}<br>
When x=2, the derivative is {{{y-8)/(y-2)}}};  it is zero when y is 8.<br>
The tangent to the graph at P(2,8) is horizontal.