# 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?

Thanks in advance for help

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?

Thanks in advance for help
Yes, use the derivative $\frac{dy}{dx} = - \frac{3x}{y}$. Suppose the two tangent points are $(x_1,y_1) \; \text{and}\; (x_2,y_2)$. Then the equation of each tangent is

$
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 $14 - y_1 = -\frac{3x_1}{y_1} \left(14 - x_1\right)$ or simplifying gives $14 y_1 + 3 \cdot 14 x_1 = 3x^1_2 + y_1^2 = 28$ or $y_1 + 3x_1 = 2$. Now eliminate $y_1$ from $3x_1^2 + y_1^2 = 28$ giving a single quadratic equation for $x_1$ (which you can solve). Using $y_1 + 3x_1 = 2$ gives you the y value.