Question 825587
<pre>
Find the slope of the tangent line:
 (a)To the ellipse x^2+y^2/2=1 at the point (1/&#8730;2,1) 
 (b)To the hyperbola x^2&#8722;y^2=1 at the point (&#8730;3,&#8730;2) 

The slope of the tangent line at a point IS the derivative
at that point.

{{{x^2+y^2/2}}}{{{""=""}}}{{{1}}}

Clear of fractions by multiplying through by 2

{{{2x^2+y^2}}}{{{""=""}}}{{{2}}}

Take the derivative term-by-term implicitly:

{{{4x*expr((dx)/(dx))+2y*expr((dy)/(dx))}}}{{{""=""}}}{{{"0"}}}

{{{expr((dx)/(dx))}}} is just 1 so

{{{4x+2y*expr((dy)/(dx))}}}{{{""=""}}}{{{"0"}}}

Subtract 4x from both sides:

{{{2y*expr((dy)/(dx))}}}{{{""=""}}}{{{-4x}}}

Divide both sides by 2y

{{{(dy)/(dx)}}}{{{""=""}}}{{{-4x/(2y)}}}

{{{(dy)/(dx)}}}{{{""=""}}}{{{-(2x)/y}}}

So we substitute the point ({{{1/sqrt(2)}}},1)

{{{(dy)/(dx)}}}@({{{1/sqrt(2)}}},1) {{{""=""}}}{{{(-2(1/sqrt(2)))/1}}}{{{""=""}}}{{{-2/sqrt(2)}}}{{{""=""}}}{{{expr(-2/sqrt(2))expr(sqrt(2)/sqrt(2))}}}{{{""=""}}}{{{-2(sqrt(2))/2}}}{{{""=""}}}{{{-sqrt(2)}}}

The other one is done the same way.

Edwin</pre>