Need help finding the tangent of an ellipse.

D

Dave J

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Dec 3, 2006
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The problem is

Find the equation of the line with a positive slope that is tangent to the ellipse
(x^2)/9 + (y^2)/4 = 1
At x=2


Now I know that to find the tangent, I find the derivative of the equation. So I got
2x/9 + 2y/4 dy/dx = 0


But its this part where I cant go any further. See I only get X, so I cant solve for dy/dx without having y in the answer, and therefor I cant solve for y.

I solved for dy/dx and got -16/18y, and just tp test I found what was Y when x = 2 for the original equation, it was 0.37.
So I adding 0.37 into -16/18y, i got -2.4 (which would be the slope of the line)

But it says find the line with the positive slope, after reading through the entire chapter in my textbook, and searching the internet, I am stuck and don't know what I'm doing wrong.
 
find the positive slope of ellipse x^2/9 +y^2/4 =1 at x=2

what is the TWO y values at x=2 ?
4/9 +y^2/4=1
y^2/4= 5/9
y^2=20/9
y=+/- 2 sqrt5/3

find the slope at 2,-2sqrt5/3

sketch the ellipse and you will see the slope is positive at this point
2x/9+2y/4 dy/dx=0
dy/dx = [-2x]4 /{ 9[2y]}
m=-8x/18y
m=-4x/9y at x=2 y=-2sqrt5/3
m=-8[3] / [-18sqrt5]
m=24/[18 sqrt5]
m=4/[3sqrt5]
m=4sqrt5/[15]

y=mx+b
-2sqrt5 /3 =4sqrt5/15 [2]+b
-10 sqrt5/15 = 8 sqrt5/15 + b
b=-2sqrt5/15

y=4sqrt5/15 x -2sqrt5/15 answer

please check for errors
Arthur
 
find the positive slope of ellipse x^2/9 +y^2/4 =1 at x=2

what is the TWO y values at x=2 ?
4/9 +y^2/4=1
y^2/4= 5/9
y^2=20/9
y=+/- 2 sqrt5/3

find the slope at 2,-2sqrt5/3

sketch the ellipse and you will see the slope is positive at this point
2x/9+2y/4 dy/dx=0
dy/dx = [-2x]4 /{ 9[2y]}
m=-8x/18y
m=-4x/9y at x=2 y=-2sqrt5/3
m=-8[3] / [-18sqrt5]
m=24/[18 sqrt5]
m=4/[3sqrt5]
m=4sqrt5/[15]

y=mx+b
-2sqrt5 /3 =4sqrt5/15 [2]+b
-10 sqrt5/15 = 8 sqrt5/15 + b
b=-2sqrt5/15

y=4sqrt5/15 x -2sqrt5/15 answer

please check for errors
Arthur
 
to solve for y, wouldn't you.

4/9 +y^2/4=1
y^2 = 5/9 = 0.555
---
4

y^2 = 0.55555
------
4

y = sqrt(0.138)

?? Thats what ive been doing and It seems to work out right, or does it not?
 
no! YOU ARE USING THE WRONG Y VALUE.
all square roots have 2 values

x^2/9 + y^2/4 =1
when x=2
4/9 +y^2/4 =1
y^2/4=5/9
y^2=4*5/9
y=+/- 2/3 sqrt5

one value of y at x=2 is 2/3 sqrt5
SECOND VALUE OF Y at x=2 is -2/3 SQRT5

USE THE SECOND Y VALUE FOR A POSITIVE SLOPE
Arthur
 
arthur ohlsten said:
no! YOU ARE USING THE WRONG Y VALUE.
all square roots have 2 values

x^2/9 + y^2/4 =1
when x=2
4/9 +y^2/4 =1
y^2/4=5/9
y^2=4*5/9
y=+/- 2/3 sqrt5

one value of y at x=2 is 2/3 sqrt5
SECOND VALUE OF Y at x=2 is -2/3 SQRT5

USE THE SECOND Y VALUE FOR A POSITIVE SLOPE
Arthur
Click to expand...

As if I looked over that, thanks for the reminder. Yhea I know what Im doing wrong now.
 
good!
If you sketch the ellipse x^2/9 + y^2/4 = 1
at y=0 x=+3 and - 3
at x=0 y=+2 and -2
it is a squashed circle

If you sketch a vertical line at x=2, it will cross the ellipse at 2 places.
at one place the tangent has a negative slope, at the other a positive slope.

I don't mean to be a pest , but sketch the ellipse to help understand the problem
Arthur
 
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