# Thread: Two tangent Lines to an ellipse

1. ## Two tangent Lines to an ellipse

From the point (14,14) there are two tangent lines to the ellipse 3x^2+y^2=28. Find the coordinates of the two points of tangency.

So I started by using implicit differentiation to find dy/dx = -3x/y and we know that y = sqrt(28-3x^2) but I'm not sure what to do from here. Would I plug in y so that dy/dx = -3x/sqrt(28-3x^2)? If that is what you do then how does that help you find the coordinates?

2. Originally Posted by Alice96
From the point (14,14) there are two tangent lines to the ellipse 3x^2+y^2=28. Find the coordinates of the two points of tangency.

So I started by using implicit differentiation to find dy/dx = -3x/y and we know that y = sqrt(28-3x^2) but I'm not sure what to do from here. Would I plug in y so that dy/dx = -3x/sqrt(28-3x^2)? If that is what you do then how does that help you find the coordinates?

Yes, use the derivative $\displaystyle \frac{dy}{dx} = - \frac{3x}{y}$. Suppose the two tangent points are $\displaystyle (x_1,y_1) \; \text{and}\; (x_2,y_2)$. Then the equation of each tangent is
$\displaystyle y - y_1 = -\frac{3x_1}{y_1} \left(x - x_1\right),\;\;\; y - y_2 = -\frac{3x_2}{y_2} \left(x - x_2\right)$
Now each tangent line passes throught (14,14) so $\displaystyle 14 - y_1 = -\frac{3x_1}{y_1} \left(14 - x_1\right)$ or simplifying gives $\displaystyle 14 y_1 + 3 \cdot 14 x_1 = 3x^1_2 + y_1^2 = 28$ or $\displaystyle y_1 + 3x_1 = 2$. Now eliminate $\displaystyle y_1$ from $\displaystyle 3x_1^2 + y_1^2 = 28$ giving a single quadratic equation for $\displaystyle x_1$ (which you can solve). Using $\displaystyle y_1 + 3x_1 = 2$ gives you the y value.