Problem: Consider an ellipse (pictured).
For any particular x-value, there are two tangents (or none).
At what x-value do the resulting tangents intersect at x = 5?
A possible solution:

The ellipse is symmetric, so finding the 1st quadrant
solution is sufficient.

We can solve for y and use the explicit function for the
top half of the ellipse without worrying about the other half. Given
𝑥2
16
+
𝑦2
8
= 1, let’s solve
for y.
Multiply through by 8 
𝑥2
2
+ 𝑦2 = 8
Subtract the stuff we don’t want. 
(this clears the fraction under the y2 term).
𝑦2 = 8 −
𝑥2
2
(we’re closer to isolating y).
Take the square root of both sides  𝑦 = ±√8 −
2𝑥 2
2
(y is by itself).
We only care about the positive arc, so we only keep the positive root: 𝑦 = √8 −

I like to name the things I’m looking for, so I’m going to say I’m looking for C, the abscissa (xvalue) of that special point of tangency that has the tangent whose x-intercept is 5.
 the point of tangency is (C, √8 −

𝑐2
).
2
Nice.
We can also find the derivative at this point. First take the derivative of y, then plug in x = c.
𝑦 = √8 −
𝑥2
2
′
so 𝑦 =
1
(8
2
−1⁄
2
𝑥2
− )
2
∙ (−𝑥) I much prefer this as a fraction: 𝑦 ′ =
We want the specific slope when x = c, so 𝑚 =
−𝑥
2
.
2√8−𝑥 ⁄2
−𝑐
2
2√8−𝑐 ⁄2
We have: a name for the x-value we want, the y-value that goes with it, and the slope of the tangent.
Put them together!
𝑥2
2
2
𝑦 − (√8 − 𝑐 ⁄2) =
−𝑐
(𝑥 − 𝑐)
2
2√8 − 𝑐 ⁄2
(
)
We want this to all work out for the point (5, 0), the target point, so let’s plug these in and simplify:
−𝑐
2
0 − (√8 − 𝑐 ⁄2) =
(
2√ 8 −
(5 − 𝑐)
𝑐2
⁄2
)
2
Multiply by 2√8 − 𝑐 ⁄2 to clear the fraction

2
−2 (8 − 𝑐 ⁄2) = (−𝑐)(5 − 𝑐)
Distribute on both sides  - 16 + c2 = -5c + c2
The c2 cancel and we’re left with c = 16/5
Is this right?!? I set up geometer’s sketchpad with the ellipse and an estimate of the tangent. Good
news! The tangent whose intercept is 5 has a point of tangency with an abscissa of x = 16/5. Whew!
f(x) = (8
0.5∙x2)0.5
xA = 3.20 yA = 1.70
5
this is the upper arch
4
these are the coordinates of the point of tangency
3
2
A
1
6
a = 0.001
4
2
2
f(xA + a) = 1.69
f(xA + a)
g(x) =
a
1
yA
∙(x
xA) + yA
this is my 'estimate' of the tangent
equation, using RDQ and x=0.001
2
3
4
5
4
6
8
10
Another approach, using implicit differentiation
The equation of the ellipse is
𝑥2
16
+
𝑦2
8
= 1 . We know that we’re going to have to work on the
derivative, so I would multiply everything by 16 and take the derivative: X2 + 2y2 = 16,
𝑑𝑦
𝑑𝑦
Which gives us 2𝑥 + 4𝑦 𝑑𝑥 = 0 or 𝑑𝑥 =
−𝑥
.
2𝑦
I want to talk about the point of tangency, so I’m just going to call the coordinates (c, b). We know from
the ellipse equation that c2 + 2b2 = 16 and that the ROC at this point is −𝑐⁄2𝑏 .
I need the equation of a tangent, so I’m going to write it using the info I have.
−𝑐
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) becomes 𝑦 − 𝑏 = 2𝑏 (𝑥 − 𝑐).
Notice the differences the variables and our names for numbers we haven’t found yet.
We know that this tangent must include the x-intercept at (5, 0), so let’s put in these values and try to
simplify. We get 0 − 𝑏 =
−𝑐
(5 −
2𝑏
𝑐) or just −𝑏 =
−𝑐
(5 −
2𝑏
𝑐)
Time for some magic. Notice that multiplying both sides by -2b would give us 2b2, which we can easily
replace from up here ®.
−𝑐
We have −𝑏 = 2𝑏 (5 − 𝑐)  multiply by -2b to get 2𝑏 2 = 𝑐(5 − 𝑐)
Which simplifies to 2𝑏 2 = 5𝑐 − 𝑐 2
Using the ellipse equation, 2b2 = 16 – c2, so we can drop it right in to get 16 − 𝑐 2 = 5𝑐 − 𝑐 2 .
The c2 terms cancel and we solve for c to get c = 16/5. BAM!
I’ve already used a picture, so let’s try some other way to support our work.
If x = 16/5, we should be able to make two tangents that intersect at x = 5. Let’s try them out and see.
The ellipse equation is
16 2
(5)
16 2
16−( )
+ 2𝑦 = 16, so 𝑦 = ±√ 2 5 , which gives us the points (3.2, 1.697),
2
with slope of about 0.943, and (3.2, -1.697) with slope of about -0.943.
This gives us tangent equations of y – 1.697 = (0.943)(x – 5) and y + 1.697 = (-0.943)(x – 5), isolating the y
lets us set them equal to each other : (0.943)(x – 5) + 1.697 = (-0.943)(x – 5) – 1.697.
1.886(x-5) = -3.394
X – 5 = -1.800, which makes x = 3.2. BAM.
The only value for x that supports the idea that the tangents meet at (5, 0) is x = 3.2, or 16/5.