I have an online homework program called Web Assign for my calculus course. It has given me this problem:
Find equations of both the tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through the point (27, 3).
First, unless I'm much mistaken, that point does not lie on the ellipse at all. Something is fishy already. What's more, unless I am mistaken yet again, ellipses always have one tangent line to every point. Regardless, I went ahead and did the problem using implicit differentiation.
2x + 18yy' = 0
18yy' = -2x
y' = -x/9y
now for the tangent line…
y = mx + b
m = -27/9*3
m = -1
3 = -27 + b
30 = b
y = 30 – x
I plugged this answer into one of the answer boxes. It says it's wrong. I tried putting DNE then undefined into both boxes, but it still said the answer was wrong. I have only one try left, so if there's something I'm missing, would somebody please tell me about it?
Best Answer
We want the tangent line(s) to pass through $(27,3)$. As you observed, this point is not on the ellipse $x^2+9y^2=81$. That's perfectly all right, but it does make the problem harder. The point $(27,3)$ is outside the ellipse, so an informal picture shows there should be two lines through $(27,3)$ that are tangent to the ellipse.
Let the point of tangency be $(a,b)$. This point is on the ellipse, and therefore $$a^2+9b^2=81.\tag{$1$}$$
By implicit differentiation, $2x+18y y'=0$. It follows that the tangent line at $(a,b)$ has slope $-a/9b$.
The equation of the line through $(a,b)$ with slope $-a/9b$ is $$y-b=-\frac{a}{9b}(x-a).\tag{$2$}$$ The point $(27,3)$ is on this line. It follows that $$27-b=-\frac{a}{9b}(3-a).\tag{$3$}$$ Simplify Equation $(3)$, and use Equation $(1)$ to find the values of $a$ and $b$. Once you have found these, you will know the equations of the required lines.