I came across the following problem from a Calculus course. Given the equation of an ellipse $9x^2+4y^2=36$ and a point $P(4,0)$. Find the equations of the two tangents to the ellipse, passing through $P$. The idea was to use the derivative, through implicit diff being $y'=\frac{-9x}{4y}$ But then I got stuck. So in order to get the answer, I tried a non calculus approach. I set up the tangent equation to be $y=m(x-4)$ and let it intersect with the ellipse by eliminating $y$. I got $9x^2+4(mx-4m)^2=36$ Removing brackets and simplifying, this becomes of quadratic nature: $$(9+4m^2)x^2-32m^2x+64m^2-36=0$$ Now an old trick I learned in High School is that in order to have tangents, the Discriminant $b^2-4ac$ must be zero. So plugging in the coefficients, I arrive at: $1024m^4-4(9+4m^2)(64m^2-36)=0$ which after simplifying becomes: $m^2=0.75$ from which I find $m=\frac{\sqrt{3}}{2}$ and $m=-\frac{\sqrt{3}}{2}$. With these slopes, the tangents are hereby found. My question: How can this be done with the derivative, provided it takes less steps than my approach? Thanks.
[Math] Tangents to an ellipse passing through an external point
algebra-precalculuscalculus
Best Answer
An alternative method would be to write the ellipse in parametric form.
Let $$x=2\cos\theta, y=3\sin \theta$$
Then the equation of the tangent is$$y-3\sin \theta=-\frac {3\cos\theta}{2\sin\theta}(x-2\cos\theta)$$
Now put $(x,y)=(4,0)$ and you get $\theta=\frac{\pi}{3},\frac{2\pi}{3}$ which leads to the equations of the tangents.