Tangents to an ellipse passing through an external point

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.

An equation for the tangent lines is ${y-y_P\over x-x_P}=y'(x,y)$. This will give you a pair of simultaneous quadratics that you can solve for $x$ and $y$. Plug these values back into $y'$.

• OK, will look into it. Thanks – imranfat Apr 19 '16 at 21:21
• @imranfat All of the second-degree terms cancel out quickly once you start solving the system. – amd Apr 20 '16 at 2:12
• Nice method, to elaborate: Let the external point be $P(x_p, y_p)$. The gradient of the line passing through $P$ is $\frac{y-y_p}{x-x_p}$. Make this equal to $\frac{dy}{dx}$, which can be found via implicit differentiation, and solve for $x$ and $y$ (These points lie on the ellipse). Plug $x$ and $y$ back into $\frac{dx}{dy}$ to find the gradient ($m$). The line equation is then: $y - y_p = m(x - x_p)$ (The $x$ and $y$ here are are just arbitrary points $x$ and $y$, don't sub in the values you got previously) – WeavingBird1917 Jun 30 '18 at 12:03

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.

• Parametric wasn't discussed at the point the question came around, but it looks much quicker, so let me ponder about it...However, if $y=0$ then $3sin\theta=0$ would give me different values of $\theta$ – imranfat Apr 19 '16 at 21:22
• I think I get it now... – imranfat Apr 19 '16 at 22:02
• There’s no need to solve for $\theta$, either. The equation of the tangents gives you a value for $\cos\theta$, from which $\sin\theta$, $x$ and $y$ are easily determined. Knowing the value of $\theta$ itself isn’t particularly illuminating to me since it isn’t the angle of the radius vector. – amd Apr 20 '16 at 2:11
• I don't get it, how is -3cosx/2sinx the slope? did you use the fact that xcos(p)/a + ysin(p) /b = 1 is always tangent or did you in fact do something geometric? – Vrisk Nov 23 '17 at 5:57
• @Vrisk the gradient is obtained by parametric differentiation – David Quinn Nov 23 '17 at 6:25