Finding the tangent of an ellipse that is perpendicular to a line The books say's 

"Find the equations of the tangents to $x^2+3y^2=4$ which are
  perpendicular to the line $x-2y=7$"

I've graphed them and found that the given line does not pass through the ellipse and the gradient of the tangent should be $dy/dx=-x/3y$. I don't know how to get an expression for the tangent because all the examples we did had a given point as well or the line intercepted the curve and i could find a point using simEqu. please help!
 A: Hint: Slope of the equation $x-2y=7$ is $0.5$. So the slope of the tangent to the ellipse is $-2$. So we get $$-2=\dfrac{-x}{3y}$$
Then find the points at which the tangent cuts the ellipse using this equation and then get the equation(s).
A: You are correct that the gradient of the tangent to the ellipse at a point $(x,y)$ will be $-x/3y$. You should then note that the equation $x - 2y = 7$ can be rearranged to $y = \frac{1}{2}x - \frac{7}{2}$ from which it is clear that the slope of the given line is $1/2$. Thus, lines perpendicular to it should have slope $-2$. This means you need tangent lines to the ellipse with slope $-2$ so we should have $-x/3y = -2$ which implies $x=6y$. Then, we need $(x,y) = (6y,y)$ to be a point on the ellipse, so it should satisfy $x^2 + 3y^2 = 4$. That is, $4 = (6y)^2 + 3y^2 = 39y^2$. So $y = \pm \sqrt{4/39}$. Then we can rearrange the ellipse equation to see that $x^2 = 4 - 3y^2 = 4 - \frac{4}{39} = \frac{152}{39}$. So points on the ellipse satisfying $-x/3y = -2$ are $(x,y) = \pm(\sqrt{4/39},\sqrt{152/39})$.
