SOLUTION: please help!!! Find the equation for the tangent lines, and the normal lines, to the hyperbolas y^2/4-x^2/2=1 when x=4.

Algebra ->  Finance -> SOLUTION: please help!!! Find the equation for the tangent lines, and the normal lines, to the hyperbolas y^2/4-x^2/2=1 when x=4.      Log On


   

Question 1080724: please help!!!
Find the equation for the tangent lines, and the normal lines, to the hyperbolas y^2/4-x^2/2=1 when x=4.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The slope of the tangent line is equal to the value of the derivative at that point.
y%5E2%2F4-x%5E2%2F2=1
y%5E2-2x%5E2=4
Implicitly differentiate,
2ydy-4xdx=0
2ydy=4xdx
dy%2Fdx=2%28x%2Fy%29
So then,
y%5E2-2%284%29%5E2=4
y%5E2-32=4
y%5E2=36
y=6 and y=-6
So for (4,6),
m=dy%2Fdx=2%284%2F6%29=4%2F3
Using the point slope form of a line,
y-6=%284%2F3%29%28x-4%29
y-6=%284%2F3%29x-16%2F3%7D%7D%0D%0A%7B%7B%7By=%284%2F3%29x-16%2F3%2B18%2F3
y=%284%2F3%29x%2B2%2F3
and for (4,-6)
m=-4%2F3
and
y%2B6=-%284%2F3%29%28x-4%29
y%2B6=-%284%2F3%29x%2B16%2F3%29
y=-%284%2F3%29x%2B16%2F3-18%2F3%29
y=-%284%2F3%29x-2%2F3
To get the normal lines, you know that the tangent and normal lines are perpendicular to each other.
So the slopes are negative reciprocals,
m%5B1%5D%2A%284%2F3%29=-1
m%5B1%5D=-3%2F4
and
m%5B2%5D%2A%28-4%2F3%29=-1
m%5B2%5D=3%2F4
So then,
(4,6)
y-6=%283%2F4%29%28x-4%29
(4,-6)
y%2B6=%283%2F4%29%28x-4%29
I leave those to you to put into slope-intercept form.
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