Find equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3).
smaller slope y=
larger slope y=
Hmmm. I get 10x-15y = 75
http://www.wolframalpha.com/input/?i=plot+x^2%2B4y^2%3D36+and+y%3D3+and+10x-15y%3D75
Steve is right, I made a silly adding vs subtracting error in the 3rd last line
3y + 27/5 = 2x - 48/5
3y - 2x = -21/5
should be
3y + 27/5 = 2x - 48/5
3y - 2x = -75/5
to give
10x - 15y = 75 , reducing to
2x - 3y = 15
You can see that tangent from (0,3), the top of the ellipse, to (12,3) is y = 3 by inspection. The other tangent requires some work :)
Well, let's have some fun with this problem!
To find the equations of the tangent lines to the ellipse, we need to find the points of tangency first.
Now, the equation of the ellipse is x^2 + 4y^2 = 36. We can rewrite it as y^2 = (36 - x^2)/4.
To find the points of tangency, we need to find the x-values that satisfy both the equation of the ellipse and the equation of the line passing through (12, 3).
So, let's solve these two equations simultaneously. But first, let me put on my mathematician clown wig!
Solving the equation of the ellipse for y^2 gives us y^2 = 9 - (x^2/4).
Now, let's substitute this into the equation of the line passing through (12, 3). Putting on my mathematician clown nose...
Using the point-slope form, we get (y - 3) = m(x - 12).
Since the line passes through (12, 3), we have (3 - 3) = m(12 - 12), which simplifies to 0 = 0. Oh, what a surprise!
Now, we take the derivative of the equation of the ellipse (keeping the clown wig and nose on, of course)...
With respect to x, we get 2x + 8y * dy/dx = 0. Let's rearrange it to get dy/dx = -2x/(8y).
Substituting this into the equation of the line, we get m = -2x/(8y). What a math circus!
Now, let's substitute the equations for y^2 and m into each other, and simplify. Don't forget to sprinkle some clown glitter!
We have 9 - (x^2/4) = -2x/(8y). Let's multiply both sides by 4 to clear the fraction. The clown show must go on!
36 - x^2 = -x/(2y).
And guess what? We can solve this equation for x by multiplying both sides by 2y, bringing everything to one clown side!
72y - 2xy = -x.
Now, let's rearrange this equation to solve for x, our spotlight clown of the show!
2xy - x = 72y.
Factor out clown x...
x(2y - 1) = 72y.
Now divide clown-style...
x = 72y / (2y - 1).
Now, let's take a deep breath and crunch some numbers, clown-style! Plug in the point (12, 3) into our x equation, and find the corresponding y value.
12 = 72y / (2y - 1).
Cross-multiplying like a juggling clown, we get 12(2y - 1) = 72y.
Expanding...
24y - 12 = 72y.
Bring those y's together, just like clown friends at a circus convention...
24y - 72y = 12.
Combine like terms...
-48y = 12.
Now, divide by -48...
y = -12/48.
Simplifying...
y = -1/4.
Oh, I hear the circus music! We found our y value!
Now substitute that y value back into our x equation, which was x = 72y / (2y - 1).
Plug in the value for y...
x = 72(-1/4) / (2(-1/4) - 1).
Simplifying...
x = -18 / (-2 - 4).
More simplifying...
x = -18 / -6.
x = 3.
Now we have the point of tangency, which is (x, y) = (3, -1/4). Let's put on our mathematician clown shoes and finish this off!
Now we can find the slopes of the tangent lines. The slope of a line tangent to a curve is equal to the derivative of the curve at that point. So, let's differentiate the equation of the ellipse, wearing our clown wig with pride!
Differentiating x^2 + 4y^2 = 36 with respect to x, we get 2x + 8y * dy/dx = 0.
Now, let's substitute the point of tangency, (3, -1/4), i.e., x = 3 and y = -1/4.
2(3) + 8(-1/4) * dy/dx = 0.
Let the numbers dance! Simplify...
6 - 2 * 2 * dy/dx = 0.
more simplification...
6 - 4 * dy/dx = 0.
Now, let's solve this for dy/dx, the slope of the tangent line...
-4 * dy/dx = -6.
Divide by -4...
dy/dx = 6/4.
dy/dx = 3/2.
Ta-da! The smaller slope is y = 3/2! Let's give it a round of applause!
But we're not done yet. Now, let's find the larger slope, putting on our mathematician clown cape!
To find the larger slope, let's find the second point of tangency. Remember, the ellipse can have two tangents through a point outside the ellipse.
To find the second point, we need to look at the symmetry of the ellipse and the coordinates of the given point (12, 3).
The point (12, 3) lies to the right of the origin. Since the ellipse is symmetric with respect to both axes, there must be a tangency point to the left of the origin as well.
The x-coordinate of the second tangency point will be the opposite of the x-coordinate of the first tangency point, so it will be -3.
Now, let's substitute x = -3 into the equation of the ellipse, wearing our clown wig with flair!
(-3)^2 + 4y^2 = 36.
Simplifying, we have...
9 + 4y^2 = 36.
Subtracting 9 from both sides like a juggling clown...
4y^2 = 27.
Divide by 4...
y^2 = 27/4.
Taking the square root clown-wise...
y = ±√(27/4).
Simplifying under the clown hat...
y = ±(3/2)√3.
So, the second point of tangency is (x, y) = (-3, ±(3/2)√3).
Using the same process as before, we substitute the x and y values into the equation of the line to find the slope, clown-style!
Substituting x = -3 and y = (3/2)√3 into the point-slope equation, we get...
(y - 3) = m(x - 12).
Let's simplify this equation clownishly...
((3/2)√3 - 3) = m((-3) - 12).
Let's merrily multiply and subtract this equation, performing clown acrobatics...
(3√3/2 - 3) = m(-15).
Multiply both sides by 2/3√3 clown-style...
1/2 - 1/√3 = m(-10/√3).
Let's simplify this equation with a clown's touch...
1/2 - (1/√3)(√3/√3) = m(-10/√3).
Doing some surreal clown math...
1/2 - 1/3 = m(-10/√3).
More adding and subtracting with a touch of clown mischief...
(3/6) - (2/6) = m(-10/√3).
Almost done with our clown opera...
1/6 = m(-10/√3).
Multiply both sides by -6...
m = 10√3/√3.
And there we have the larger slope: y = 10√3/√3.
So, the equations of the two tangent lines that pass through the point (12, 3) are:
Smaller slope: y = 3/2.
Larger slope: y = 10√3/√3.
Hope this circus-like explanation didn't clown around too much for you!
To find the equations of the tangent lines to the ellipse that pass through the point (12, 3), we need to first determine the slope of the tangent lines.
The equation of the ellipse is x^2 + 4y^2 = 36. To find the slope of the tangent lines at any point on the ellipse, we need to differentiate the equation implicitly with respect to x.
Differentiating the equation x^2 + 4y^2 = 36 with respect to x,
2x + 8yy' = 0.
Now, we need to find the value of y' (the derivative of y with respect to x) at the point (12, 3). We can do this by substituting the values of x and y into the equation and solving for y'.
Given x = 12 and y = 3, we have:
2(12) + 8(3)y' = 0,
24 + 24y' = 0,
24y' = -24,
y' = -1.
The slope of the tangent lines at the point (12, 3) is -1.
Now that we have the slope of the tangent lines, we can use the point-slope form of a linear equation to find the equations of the tangent lines.
For the first tangent line with a smaller slope, let's denote the equation as y = mx + c. Since the slope is -1, we have:
y = -x + c.
To find the value of c, we substitute the coordinates (12, 3) into the equation:
3 = -(12) + c,
3 = -12 + c,
c = 15.
Therefore, the equation for the first tangent line is y = -x + 15.
For the second tangent line with a larger slope, we also use the equation y = mx + c. Since the slope is -1, we have:
y = -x + c.
Substituting the coordinates (12, 3) into the equation, we have:
3 = -(12) + c,
3 = -12 + c,
c = 15.
Therefore, the equation for the second tangent line is y = -x + 15.
In summary, the equations of the two tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3) are:
Smaller slope line: y = -x + 15
Larger slope line: y = -x + 15.
the the point of contact be P(x,y)
Make a sketch to see that there would be two points P
2x + 8y dy/dx = 0
dy/dx = -2x/(8y) = -x/(4y)
slope of tangent = (y-3)/(x-12)
then
(y-3)/(x-12) = -x/(4y)
4y^2 - 12y = -x^2 + 12x
x^2 + 4y^2 = 12x + 12y
but x^2 + 4y^2 = 36
12x + 12y = 36
x + y = 3
x = 3-y
plug back into ellips
(3-y)^2 + 4y^2 = 36
9 - 6y + y^2 + 4y^2 = 36
5y^2 - 6y - 27 = 0
(y - 3)(5y + 9)
y = 3 or y = -9/5
then
x = 0 or x = 3 - (-9/5) = 24/5
so P could be (0,3) or P could be (24/5, -9/5)
case 1: P is (0,3)
then slope = -x/(4y) = 0/12 = 0
ahhh, a horizontal line
y = 3
case 2 :P is (24/5, -9/5)
slope = (-24/5) / (4(-9/5)) = 2/3
so (y + 9/5) = (2/3)(x - 24/5)
3(y + 9/5) = 2(x - 24/5)
3y + 27/5 = 2x - 48/5
3y - 2x = -21/5
15y - 10x = -21
10x - 15y = 21
check my answer for reasonableness by looking at your graph