Okay, so I've been grinding away at this problem for a few hours and I cannot figure out what I am supposed to do. I am given on my worksheet that I need to find the correct value for a and b for the equation in the form x^2 + xy + ay^2 = b. (It's a tilted ellipse.) There is a point on the graph at (1,3) and the tangent line at that point has a slope of -5/14.
Okay... so what am I doing wrong?
At first, I tried plugging values in for a and b and eventually I found I a pattern; the closest I got to matching the information was with the equation x^2+xy+2.055y^2=24.55. By implicit differentiation, the derivative would be -(100(2x+y))/(100x+411y), and after plugging in (1,3) I was soooo close.
After that method got me nowhere as I was testing and testing different values, I even tried the (y2-y1)/(x2-x1)=the slope (which I used as a, then solved for b) formula, which I had a feeling wouldn't get me anywhere since it supposed to be for lines. It didn't. (The graph I got didn't even have the point (1,3) on it.)
So basically I've been plugging in values for a and b I know are close, have the point (1,3) on the graph, calculating the implicit derivative, and NOT ever getting the tangent slope for that point, which has to be exactly -5/14.
I know there has to be an easier way to do this... but I was never taught how to solve this problem...
If anyone has a method for me to use, please do tell!
Okay... so what am I doing wrong?
At first, I tried plugging values in for a and b and eventually I found I a pattern; the closest I got to matching the information was with the equation x^2+xy+2.055y^2=24.55. By implicit differentiation, the derivative would be -(100(2x+y))/(100x+411y), and after plugging in (1,3) I was soooo close.
After that method got me nowhere as I was testing and testing different values, I even tried the (y2-y1)/(x2-x1)=the slope (which I used as a, then solved for b) formula, which I had a feeling wouldn't get me anywhere since it supposed to be for lines. It didn't. (The graph I got didn't even have the point (1,3) on it.)
So basically I've been plugging in values for a and b I know are close, have the point (1,3) on the graph, calculating the implicit derivative, and NOT ever getting the tangent slope for that point, which has to be exactly -5/14.
I know there has to be an easier way to do this... but I was never taught how to solve this problem...
If anyone has a method for me to use, please do tell!
-
x² + xy + ay² = b
I'll have to repeat some work that you've done already.
2x + y + xy' + 2ayy' = 0
y' = -(2x + y)/(x + 2ay)
We're told that at (1, 3), y' = -5/14
therefore
5/14 = (2 + 3)/(1 + 6a)
1 + 6a = 14
a = 13/6
Then, since (1, 3) satisfies the equation of the ellipse, we have
1 + 3 + 9a = b and since a = 13/6, 9a = 3*13/2
4 + 39/2 = b
i.e. b = 47/2
Those values you found aren't very far away from these.
I'll have to repeat some work that you've done already.
2x + y + xy' + 2ayy' = 0
y' = -(2x + y)/(x + 2ay)
We're told that at (1, 3), y' = -5/14
therefore
5/14 = (2 + 3)/(1 + 6a)
1 + 6a = 14
a = 13/6
Then, since (1, 3) satisfies the equation of the ellipse, we have
1 + 3 + 9a = b and since a = 13/6, 9a = 3*13/2
4 + 39/2 = b
i.e. b = 47/2
Those values you found aren't very far away from these.