# line tangent to an elipse given one point

• Apr 12th 2005, 06:48 PM
bdj
line tangent to an elipse given one point
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
• Apr 12th 2005, 10:08 PM
paultwang
1. Derive y1(x) and y2(x) for the top and bottom parts of the eclipse.

2. The derivative of y1(x) at the tangent is the same as the slope of the tangent line.
• Apr 13th 2005, 03:16 AM
ticbol
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)

------------------
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.]

-------------------
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.