Write the equations of the tangent lines from (3,0) to x^2 + 2y^2 = 6
Anyone have any idea? I've graphed it but i still have no clue where it's tangent to the elipse.
-bdj
Here is one way.
The slope of the tangent line is the 1st dervative, the dy/dx or y'.
x^2 +2y^2 = 6 ....(1)
Differentiate implicitly with respect to x,
2x +4y*y' = 0
4y' = -2x
y' = -2x/(4y) = -x/(2y).
Let (x,y) be the point of tangency.
So we have two points on the tangent line: (3,0) and (x,y).
We can get the slope of the tangent line by
slope, m = (y-0)/(x-3) = y/(x-3).
This m is the same as y', so,
-x/(2y) = y/(x-3)
Cross multiply,
-x(x-3) = (2y)y
-x^2 +3x = 2y^2 .....(2)
From (1),
x^2 +2y^2 = 6
2y^2 = 6 -x^2
Substitute that into (2),
-x^2 +3x = 6 -x^2
3x = 6
x = 6/3 = 2 ...***
Substitute that into (1),
x^2 +2y62 = 6
2^2 +2y^2 = 6
2y^2 = 6 -2^2 = 2
y^2 = 2/2 = 1
y = +,-sqrt(1)
y = 1 or -1.
That means there are 2 points of tangency: (2,1) and (2,-1)
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So, for point (2,1),
Equation of tangent line from (3,0) is
y' = m = -x/(2y) = -2/(2*1) = -1
By point-slope form of the equation of a line,
(y -y1) = m(x -x1)
Using (2,1) as (x1,y1),
(y-1) = -1(x -2)
y -1 = -x +2
x +y -1 -2 = 0
x +y -3 = 0 ....the tangent line.
[if using (3,0) as (x1,y1),
(y -0) = -1(x -3)
y = -x +3
x +y -3 = 0 ....the tangent line, same.]
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For point (2,-1)
m = -x/(2y) = -2/(2* -1) = -2/(-2) = 1
(y -y1) = m(x -x1)
(y -(-1)) = 1(x -2)
y +1 = x -2
y -x +1 +2 = 0
x -y -3 = 0 ...the tangent line.