# Finding tangent lines to an ellipse that pass through a given point

1. Nov 19, 2008

### SolidSnake

1. The problem statement, all variables and given/known data
Find the equations of all the tangent lines to x^2 + 4y^2 = 36 that pass through the point (12,3)

2. Relevant equations
the derivative of the ellipse is dy/dx = -2x/8y
(I'm not sure if that is correct, i have only recently learned implicit differentiation.)

3. The attempt at a solution
Using a point on the ellipse as ( x , +- ((6 - x) /2)) and the point which was given, i used the slope formula for a line and then set that equal to the derivative. However, now i'm left with 2 variables, x and y. I am 100% positive that we aren't suppose to use a CAS to solve this question. Could someone please help me out

2. Nov 19, 2008

### Staff: Mentor

Your equation for y at a point on the ellipse is wrong. The square root of (36 - x^2)/4 is not equal to (6 - x)/2.

I drew a quick sketch of the ellipse and found that its vertices are at (0, +/-3) and at (+/-6, 0). One of the tangent lines is horizontal.

3. Nov 19, 2008

### SolidSnake

then a point on the ellipse would be (x , root of (36 - x^2)/4 ) correct?. but even then wouldn't that leave me with the same problem when i set m = m

4. Nov 19, 2008

### Staff: Mentor

Any point on the ellipse is (x, +/-sqrt((36 - x^2)/4).

The equation of the line is
$$-\frac{\sqrt{36 - x^2}}{2} - 3 = \frac{x}{2\sqrt{36 - x^2}}(x - 12)$$
For the first expression on the left side, I chose the negative root, since I want a negative y value at the point of tangency in the fourth quadrant.
Multiply both sides by the expression in the denominator on the right side.
$$-(36 - x^2) - 6\sqrt{36 - x^2} = x^2 - 12x$$
Now move everything but the radical to the other side.
$$- 6\sqrt{36 - x^2} = - 12x + 36$$
Simplify a bit.
$$\sqrt{36 - x^2} = 2x - 6$$
Square both sides.
$$36 - x^2 = 4x^2 - 24x + 36$$
$$5x^2 - 24x = 0$$
The last expression on the left side can be factored to give the x-values that are on the line, and that have the right slope. As it turned out, I didn't need to be concerned about using a negative expression for the y-value in the 4th quadrant, because I squared both sides later on. You do need to take into account that the y-value is negative at one point of tangency, though.