<li>find the lines that are a. tangent and b. normal to the curve</li>
</ol>
<p>x^2 + xy - y^2 = c at the pt (2,3)</p>
<p>thanks so much</p>
<p>bump.....please someone</p>
<p>I haven't done implicit differentiation since like last Autumn, so I could be off, but here's my best shot as of this post...</p>
<p>Step 1. differentiate both sides of the equation...
2x + x(dy/dx) + y(1) - 2y(dy/dx) = 0</p>
<ol>
<li><p>Collect dy/dx terms, and factor...
dy/dx(x - 2y) = -y - 2x</p></li>
<li><p>Solve for dy/dx...</p></li>
</ol>
<p>dy/dx = (-y - 2x)/(x - 2y)</p>
<ol>
<li>Evaluate this expression at (2, 3)...</li>
</ol>
<p>dy/dx |(2, 3) = (-3 - 2<em>2)/(2 - 2</em>3) = (-7/-4) = 7/4. This is the slope of the tangent line at (2, 3) of the curve.</p>
<ol>
<li>Use the point-slope formula to find the equation of the tangent line...</li>
</ol>
<p>y - 3 = 7/4(x - 2)
y = (7/4)x - (7/2) + 3
a) y = (7/4)x - (1/2)</p>
<ol>
<li>Now, all we have to do for b is repeat the point-slope formula with the normal line's slope, which is -4/7.</li>
</ol>
<p>y - 3 = -4/7(x -2)
y = -(4/7)x + (8/7) + 3
b) y = -(4/7)x + (29/7)</p>
<p>Hope this helps. Good luck! :)</p>
<p>thanks! yes i understand it now</p>