Vryyyy hard calc question? Help!?

Question

Find the equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12,3).

y = (horizontal tangent line)
y = (non-horizontal tangent line)

any ideas?

Ok but wait....15 is the y intercept. But it wants the eq. in the form y=...

so there cant be any y's in the equation..right?

Answer 1

First off, rewrite the equation of the ellipse as:

x^2 + (2y)^2 = 6^2.

Rewriting it this way will allow us to find the intercepts much easier. The general (simplified) formula of an ellipse is:

(ax)^2 + (by)^2 = r^2, where r is the radius, a is the coefficient in front of x, and b is the coefficient in front of y. In this case:

a=1, b=2 and r = 6.

So what does this mean? The x-intercepts are always (in cases where the 'center' of an ellipse is located at the origin, (0,0)) located at +/- r/a, whereas the y-intercepts are always located at +/- r/b.

So: There are x-intercepts at x = {6, -6} and y-intercepts at y= {3, -3}.

We need the tangent lines to cross at (12,3). However, one thing which is apparent (once you draw the ellipse and the point) is that the tangent line at point (0,3) will cross the point (12,3) since they both share the same y-coordinate, and the tangent line at (0,3) has a slope of 0.

Thus: the equation of the Horizontal Tangent Line is y=3.

To get the equation of the second tangent line that passes through point (12,3) is a lot trickier. First, rewrite the equation of the ellipse.

1) x^2 +4y^2 = 36
[given]
2) 4y^2 = 36 - x^2
[subtract x^2]
3) y^2 = 9 - x^2 /4
[divide by 4]
4) y = +/- Sqrt(9 - x^2 /4)
[take the square root of both sides]

However, we can further simplify this equation! If we draw out the ellipse, we can see the other tangent line that passes point (12,3) MUST come from the 'bottom half' of the eliipse (the parts under the x-axis). Because of this, we only need to focus on the negative 'part' of the ellipse equation. Thus:

Let y = - Sqrt(9 - x^2 /4).

In order to go further, we must understand what we're looking for. The slope of the ellipse is constantly changing, as x changes. However, the slope of the tangent line is always constant, since the tangent line is a straight line. Because of this, we can derive two equations which we can use to find x, and y.

We know that the tangent line must have a slope of:

Slope = (y - 3) / (x - 12), where x, and y are the coordinates of a point on the ellipse. (this is basically just the slope formula for straight lines, except I already put the point (12,3) in).

Now, in order to find the slope of the ellipse at point (x,y), we must take derivatives relative to x. Thus:

1) y = - Sqrt(9 - x^2 /4)
[derived from previous part]
2) y' = (1/2) * (9 - x^2 /4)^(-1/2) * (1/2) * x
[First order derivatives. Note that "*" is just the multiplication sign]
3) y' = x / [4Sqrt(9 - x^2 /4)]
[Rewriting and simplifying gives us this equation]

This is the derivative (or slope) of the ellipse at point (x,y), because y' is essentially another name for slope.

Thus, in order for a tangent line to be tangent to a point (x,y) AND cross the point (12,3), it MUST meet one requirement.

Slope = (y - 3) / (x - 12) = y' = x / [4Sqrt(9 - x^2 /4)]

So:

1) (y - 3) / (x - 12) = x / [4Sqrt(9 - x^2 /4)]
[derived from previous parts]
2) y - 3 = [x^2 - 12x] / [4Sqrt(9 - x^2 /4)]
[multiply by x -12]
3) y = [x^2 - 12x] / [4Sqrt(9 - x^2 /4)] + 3
[add 3 to both sides]
4) - Sqrt (9 - x^2 /4) = y = [x^2 - 12x] / [4Sqrt(9 - x^2 /4)] + 3
[We substituted our original equation for y back into the equation here! Doing this allows us to eliminate the variable y, leaving us only one variable to work with!]
5) - Sqrt (9 - x^2 /4) = [x^2 - 12x] / [4Sqrt(9 - x^2 /4)] + 3[4Sqrt(9 - x^2 /4)] / [4Sqrt(9 - x^2 /4)]
[Find a common denominator between the two terms on the right hand side, in this case, [4Sqrt(9 - x^2 /4)])
6) - Sqrt (9 - x^2 /4) =[x^2 - 12x +3[4Sqrt(9 - x^2 /4)] ] / [4Sqrt(9 - x^2 /4)]
[Please bear with me, this part is ugly to write out in text, I simply combined the two terms, since they both had the same denominator]
7) -(36 - x^2) = x^2 - 12x + 12Sqrt(9 - x^2 /4)
[Multiply both sides by 4Sqrt(9 - x^2 /4)]
8) x^2 - 36 = x^2 -12x +12Sqrt(9 - x^2 /4)
[Rearrange left hand side]
9) 12x - 36 = 12Sqrt(9- x^2 /4)
[Subtract x^2 from both sides, and add 12x to both sides]
10) x - 3 = Sqrt(9 - x^2 /4)
[Divide both sides by 12]
11) x^2 - 6x + 9 = 9 - x^2 /4
[Square both sides to get rid of the root]
12) x^2 - 6x = -x^2 /4
[Subtract 9 from both sides]
13) x - 6 = -x /4
[Divide both sides by x]
14) 5x = 24
[Add x/4 to both sides, then add 6 to both sides, then multiply both sides by 4]
15) x = 24/5 = 4.8

We found earlier that y = - Sqrt (9 - x^2 /4)

Plugging in x gives us: y = -1.8.

Thus, (x, y) = (4.8, -1.8).

But, how can we be sure that this is the correct answer? The simplest way to double check is to check two things.
1) Does this point exist on the ellipse?
By plugging x and y back into the original equation (x^2 + 4y^2 = 36), we find that

(4.8)^2 + 4(-1.8)^2 = 36.

Thus, the point (4.8, -1.8) does exist on the ellipse.

2) The second thing which we must confirm is whether or not the tangent line at this point really passes through (12,3).

Earlier, we said that if it did pass through (12,3), the slope of the tangent line must equal the derivative at the point (x, y).

By plugging in x and y back into the slope formula, we find that:

Slope = 2 / 3 = .6666667

By plugging in x into our first order derivatives, we also get the result that:

y' = 2 / 3 = .6666667

Thus, we can conclude that the tangent line at point (4.8, -1.8) really does pass through the point (12, 3). Furthermore, the slope of this tangent line is 2/3.

Thus, the equation of the non-horizontal line is:

(y - 3) = (2/3)* (x - 12) [We can get this using the point-slope formula]

If you want the answer in the form of y = mx + b, you get:

Non-horizontal Tangent Line = y = 2x /3 - 5.

Thus: to summarize:

y = 3
y = 2x /3 - 5.

These are the equations of the two tangent lines!

Hope this helps.

Edit 1: To the Asker:

The y-intercept is not 15. I suggest you simply ignore the first 2 posters since they are wrong (don't let them mislead you). As of now, the only people who have posted correct answers are Scarlet (the poster above me) and myself. He used implicit differentiation to solve it (the short, more advanced, way), while I used regular differentiation to solve it (the long way). Both ways work. Hope this helps, and good luck with the rest of your problem set.

Answer 2

The equation for the horizontal tangent line is easy:
m = 0 and it passes through (12,3), so
y - 3 = 0(x - 12)
y = 3

To verify that it is tangent to the ellipse,
x^2 + 4(3^2) = 36
x^2 + 36 = 36
x^2 = 0
x = 0

Since there is only ONE such solution, it touches the ellipse at just ONE point, and is therefore tangent to the ellipse at that point.


The second part, asking for the non-horizontal line is the challenge here. It's slope must match the slope of the ellipse AND pass through (12,3). By drawing the picture, it is obvious that the other point is in the 4th quadrant. So when solving for y (to then find the slope), we want the negative square root.
x^2 + 4y^2 = 36
4y^2 = 36 - x^2
y^2 = 9 - (1/4)x^2
y = -sqrt[9 - (1/4)x^2]

dy/dx = 1/4(2x)[9 - (1/4)x^2]^-1/2
dy/dx = x/sqrt[9 - (1/4)x^2] = slope

Also, m = (y - 3)/(x - 12)
= { -sqrt[9 - (1/4)x^2] - 3}/(x - 12)

Setting these equal to each other, I get a mess:
{ -sqrt[9 - (1/4)x^2] - 3}/(x - 12) = x/sqrt[9 - (1/4)x^2]
sqrt[9 - (1/4)x^2]{-sqrt[9 - (1/4)x^2] - 3} = (x-12)x
-9 + (1/4)x^2 - 3sqrt[9 - (1/4)x^2] = x^2 -12x
3sqrt[9 - (1/4)x^2] = -(3/4)x^2 + 12x -9
sqrt[9 - (1/4)x^2] = -(1/4)x^2 + 4x - 3
9 - 0.25x^2= 0.0625x^4 - 2x^3 + 17.5x^2 - 24x + 9
0 = 0.0625x^4 - 2x^3 + 17.75x^2 - 24x

x = 0 (which we already know) or

0 = 0.0625x^3 - 2x^2 + 17.75x - 24
0 = x^3 - 32x^2 + 280x - 384
Have I made a mistake in my algebra? I'm not seeing how to get an answer from here.

Answer 3

(Note to kellenraid: (12, 3) is not a point on the ellipse, it is a point on the tangent line. If it was on the ellipse it would be easy!)

Implicit differentiation gives
2x + 8y dy/dx = 0
so when y = 0 (and x = ± 6) the tangents are vertical, otherwise dy/dx = -x / (4y). Obviously the vertical tangents don't pass through (12, 3)!
At a point (a, b) on the ellipse, the equation of the tangent line is therefore
y - b = (-a / (4b)) (x - a)
For this to pass through (12, 3) we must have
3 - b = (-a / 4b) (12 - a)
and multiplying through by 4b gives
12b - 4b^2 = -12a + a^2
or a^2 + 4b^2 - 12a - 12b = 0.
Also, since (a, b) is on the ellipse, we know
a^2 + 4b^2 = 36
so we must have 12a + 12b = 36 <=> a + b = 3.
Set b = 3-a, then a^2 + 4(3-a)^2 = 36
<=> 5a^2 - 24a = 0
<=> a = 0 or a = 24/5.

a = 0 gives b = 3 and the line y = 3.

a = 24/5 gives b = 3 - 24/5 = -9/5 and the line
y - b = (-a / (4b)) (x - a)
<=> y + 9/5 = [(-24/5) / (-36/5)] (x - 24/5)
<=> y + 9/5 = (2/3) x - 16/5
<=> y = (2/3) x - 5.

Answer 4

Y = 3 is the horizontal line. You can see this by graphing. I believe y = (2/3)x - 5 is the other tangent. I can only find the one intersection on the TI-84, but I do not have a non-graphical solution.

Answer 5

First solve for y
4y^2 = 36 - x^2
y^2 = 9 - (1/4)x^2
Take the derivative with respect to x of both sides:
2y dy/dx = 0 - 2(1/4)x= -(1/2)x
dy/dx = -(1/4)(x/y)
The slope-intercept form of a straight line is
f(x) = mx + b where m is the slope and b the y-intercept
m in this case is dy/dx, so
f(x) = -(1/4)(x^2/y) + b
f(x=12) = -(1/4)(144/3) + b = 3
= -(144/12) + b = 3
= -12 + b = 3
b = 3 + 12 = 15
The rest you should be able to do easily

Answer 6

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