Question 1016362
Differentiate,
{{{(2xdx)/81+(2ydy)/64=0}}}
{{{(ydy)/64=-(xdx)/81}}}
{{{dy/dx=-(x/y)(64/81)}}}
The slope of the tangent line is equal to the value of the derivative.
An angle of 45 degrees is equivalent to a slope of 1.
{{{1=-(x/y)(81/64)}}}
{{{y=-(64/81)x}}}
This point also satisfies the ellipse equation,
{{{x^2/81+(1/64)(64/81)^2x^2=1}}}
{{{(1/81+64/81^2)x^2=1}}}
{{{(145/6561)x^2=1}}}
{{{x^2=6561/145}}}
{{{x=0 +- 81/sqrt(145)}}}
{{{x=0 +- (81/145)sqrt(145)}}}
Then plugging that into the ellipse equation you get,
{{{y= 0 +- (64)/sqrt(145)}}}
{{{y=0 +- (64/145)sqrt(145)}}}
So the two points are,
({{{(81/145)sqrt(145)}}},{{{-(64/145)sqrt(145)}}})
({{{-(81/145)sqrt(145)}}},{{{(64/145)sqrt(145)}}})
That's when the slope is 1.
Similarly when the slope is -1.
({{{(81/145)sqrt(145)}}},{{{(64/145)sqrt(145)}}})
({{{-(81/145)sqrt(145)}}},{{{-(64/145)sqrt(145)}}})
Use the point slope form of a line to get the equation of the tangent line.
{{{y+64/sqrt(145)=1(x-81/sqrt(145))}}}
{{{y=x-145/sqrt(145)}}}
{{{y=x-sqrt(145)}}}
Similarly,
{{{y=x+sqrt(145)}}}
{{{y=-x+sqrt(145)}}}
{{{y=-x-sqrt(145)}}}
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*[illustration dc23.JPG].