x^2 + 2x + 4y^2 - 16y + 13v = 0 represents an ellipse
the line y=mx meets the ellipse
show that 3m^2 - 4m - 3 => 0.
What value(s) of m make the line a tangent to the ellipse?
1. I assume that you mean:
$\displaystyle x^2 + 2x + 4y^2 - 16y + 13 = 0$
2. Calculate the points of intersection between the line and the ellipse substituting y = mx:
$\displaystyle x^2 + 2x + 4m^2 x^2 - 16 m x + 13 = 0$ and solve for x. (Use the formula to solve a quadratice equation)
$\displaystyle x = - \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} - 8m + 1}{4m^2 + 1} \vee x = \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} + 8m - 1}{4m^2 + 1} $
3. You only get one point of intersection = tangent point if the radicand equals zero.
4. To calculate the values of m, solve for m:
$\displaystyle 3m^2 - 4m - 3 = 0$ (Use the formula to solve a quadratice equation)