given the curve x + xy + 2y^2 = 6...
a. find an expression for the slope of the curve.
i got (-1-y)/(x + 4y) as my answer.
b. write an equation for the line tangent to the curve at the point (2,1).
i got y = (-1/3)x + (5/3).
but i didn't any answer for c!
c. find the coordinates of all other points on this curve with slopes equal to the the slope at (2,1).
do i just set (-1-y)/(x+4y) equal to -1/3?
a) is correct
b) correct
c) yes, set (-1-y)/(x+4y) equal to -1/3
solving this I got x = 3-y
sub that back into the original equation to get after simplifying
y^2 + 2y - 3 = 0
(y+3)(y-1) = 0
y = -3 or y = 1
but those back into x=3-y for
x =6 or x = 4
so the other points are (6,-3) and (4,1)
you better check my math, it is getting late here.
the second one would be x = 2, resulting in the original point (2,1), i think. but other than that, thank you sooooo much!
i mean: do i just set (-1-y)/(x+4y) equal to -1/3 AND use the original formula to come up with an answer? like a two-variable system?
ALSO ALSO ALSO - on my first calc question (a couple posts down!), i added more to the problem that i forgot, so i need help with that too.
Well, if you want to find the coordinates of other points on the curve with the same slope as at (2,1), you would need to set the expression for the slope equal to -1/3 and solve for x and y.
So, we start with (-1-y)/(x+4y) = -1/3. Let's cross multiply to get rid of the fractions:
3(-1-y) = -1(x+4y)
Simplifying that:
-3-3y = -x-4y
Next, we can rearrange the terms to isolate x:
-4y + 3y = -x - 3
Simplifying again:
-y = -x - 3
Now, let's rearrange the equation to solve for y:
y = x + 3
Now we have an equation for the curve that gives us all the points with the same slope as at (2,1). So, the coordinates of these points can be written as (x, x+3).
I hope that clears things up. Happy clowning around with math!
To find the coordinates of all other points on the curve with slopes equal to the slope at (2,1), you need to set the expression for the slope (-1-y)/(x + 4y) equal to the slope at (2,1), which is -1/3.
So the equation becomes: (-1-y)/(x + 4y) = -1/3.
To solve for the coordinates of those points, you need to simultaneously solve this equation along with the original equation of the curve, which is x + xy + 2y^2 = 6.
Here's how you can solve for the coordinates of other points:
1. Start by plugging in the coordinates of the point (2,1) into the equation of the curve:
2 + 2(1) + 2(1)^2 = 6
Simplifying, we get 2 + 2 + 2 = 6
6 = 6
Since the equation holds true, the point (2,1) is on the curve.
2. Next, substitute the value of x in the equation (-1-y)/(x + 4y) = -1/3 as 2, and the value of y as 1:
(-1-1)/(2 + 4(1)) = -1/3
-2/6 = -1/3
Simplifying, we get -1/3 = -1/3
Again, the equation holds true, meaning the slope at (2,1) satisfies the equation.
Now, to find the coordinates of all other points with the same slope, you need to eliminate y from the system of equations.
3. Start by rearranging the equation of the curve, x + xy + 2y^2 = 6:
xy + 2y^2 = 6 - x
4. Multiply both sides of the equation by 3 to eliminate the denominators in the equation (-1-y)/(x + 4y) = -1/3:
-3 - 3y = x + 4y
5. Rearrange the equation to solve for x:
x = -3 - 3y - 4y
x = -3 - 7y
6. Substitute this expression for x in the rearranged curve equation:
(-3 - 7y)y + 2y^2 = 6 - (-3 - 7y)
Simplify:
-3y - 7y^2 + 2y^2 = 6 + 3 + 7y
-7y^2 - y = 9 + 7y
Rearrange and combine like terms:
7y^2 + 8y - 9 = 0
7. Now you have a quadratic equation. Solve it by factoring, completing the square or using the quadratic formula.
Once you find the values of y that satisfy the equation, substitute them back into the equation x = -3 - 7y to get the corresponding x-values. Each pair of (x, y) values you obtain will represent a point on the curve with the same slope as (2,1).
Note: The equation you derived for the slope and the equation for the line tangent to the curve at (2,1) seem to be correct based on your calculations for parts a and b.