SOLUTION: please help me to solve this problem: Find the equation of the tangent to the hyperbola x^2-4y^2=36 which is perpendicular to the line x-y+4=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: please help me to solve this problem: Find the equation of the tangent to the hyperbola x^2-4y^2=36 which is perpendicular to the line x-y+4=0       Log On


   

Question 983166: please help me to solve this problem:
Find the equation of the tangent to the hyperbola x^2-4y^2=36 which is perpendicular to the line x-y+4=0
Thanks a lot
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
A line perpendicular to x-y%2B4=0 would have a slope that is the negative reciprocal of its slope.
y=x%2B4
m=1
m%5Bp%5D=-1
So the slope of the perpendicular line would be m=-1.
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The tangent line to the hyperbola can be found by differentiating,
2x-8y=0
2xdx=8ydy
m=dy%2Fdx=%282x%29%2F%288y%29=x%2F%284y%29
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This slope must equal the previously calculated slope,
x%2F%284y%29=-1
4y=-x

x%5E2%281-1%2F4%29=36
%283%2F4%29x%5E2=36
x%5E2=48
x=0+%2B-+4sqrt%283%29
So then,
y=0+%2B-+sqrt%283%29
The two points of intersection are,
(-4sqrt%283%29,sqrt%283%29) and (4sqrt%283%29,-sqrt%283%29)
So then using the point-slope form of a line,
y-sqrt%283%29=-%28x%2B4sqrt%283%29%29 and y%2Bsqrt%283%29=-%28x-4sqrt%283%29%29
highlight_green%28y=-x%2B3sqrt%283%29%29 and highlight_green%28y=-x-3sqrt%283%29%29
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