# Possibly flawed calculus homework question (tangent to ellipse)

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?

• Could the problem have a typo? If the equation had $810$ rather than $81$, at least the point $(27,3)$ would be on the ellipse. Nov 5, 2012 at 3:37
• Your methods are correct. It would appear the problem is wrong Nov 5, 2012 at 3:37
• I'm tending to agree with Jak... Nov 5, 2012 at 3:40
• My interpretation of the problem is this: given a point that is outside the ellipse, there are two lines that are tangent to the ellipse that pass through that point.
– wj32
Nov 5, 2012 at 3:42
• @BigEndian, try wj32's interpretation. Check math.stackexchange.com/questions/58055/… if you need some help Nov 5, 2012 at 3:44

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.

• Ah! I completely misread the problem. They're not giving me the point of tangency, they're giving me a point on the curve. I can solve this now... Nov 5, 2012 at 3:54
• Well, they are not giving you a point on the curve, they are giving you a point outside the curve. Nov 5, 2012 at 3:56
• On the... Whatever... Yes I see what you mean. But the idea is the same, I was reading the question wrong. Nov 6, 2012 at 7:33

The problem is correct. There are two distinct tangent lines to the ellipse which pass through the point. The point is not on the ellipse (otherwise, yes, there would be only one). Figuring out the points at which those lines are tangent to the ellipse is part of the problem.

• counbtinghaus:I agree, and was just going to write that. I recall doing many a problem with as simple a curve as $y=x^2$, and given a partricular point such as $(5,4)$ which is outside of the convex hull of the curve, to find the two tangent lines. Nov 5, 2012 at 3:47