I was revising for one of my end of year maths exams, then I came across this example on how to find lines of tangents to ellipses outside the curve. Personally, I'd use differentiation and slopes to find such lines, but the lecturer does something simpler and more elegant.
The question is: "Find the equations of the lines through (1, 4) which are tangents to the ellipse $x^2 + 2y^2 = 6$"
And then we put the lines into the standard form, which comes out as $ y = mx − (m − 4)$ where $m$ is the slope of the line.
Then, the lecturer substitutes the equation we got into the original equation of the curve, which we get $x^2 + 2[mx − (m − 4)]^2 = 6$.
Now, the lecturer goes from the equation above to something I can't understand how to derive. With the explanation "We now look for repeated roots in the equation, as each tangent meets the line exactly once, we get":
$[4m(m − 4)]^2 − 4(1 + 2m^2 )(2m^2 − 16m + 26) = 0$
Can you guys please help me understand how to get to the equation above? I tried using the quadratic formula, where x has repeated roots (in other words, the rational bit is zero), but I still got something entirely different.
Thanks.
A line is tangent to a conic if there is just one point of intersection, i.e. if the last equation has two identical solutions, that is the same as requesting that the discriminant of the quadratic equation is zero.
By the way, you can also use a homotethy $\varphi$ bringing the ellipse and the exterior point in a circle and a exterior point, solve this simpler problem, then apply $\varphi^{-1}$.