tangent lines...?

Question

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

there's a horizontal and non-horizontal tangent line...and i have no clue how to find it...

Answer 1

The horizontal one is simply y = 3

Let the tangent touch the ellipse at (p,q)
where p^2 + 4q^2 = 36
It can be seen from a rough sketch that the y cordinate, q < 0

The slope of the tangent = slope of ellipse at point in question
Differentiating equation of ellipse
2x + 8y*dy/dx = 0
dy/dx = -2p / 8q = -p / (4q)
4q LOL

Also the equation of the line is
(q - 3) / (p - 12) = -p / 4q
4q^2 - 12q = -p^2 + 12p
Substituting p^2 + 4q^2 = 36,
we get q = 3 - p

So 4q^2 + p^2 = 36
5p^2 - 24p = 0
p = 0, 24/5
p = 0 corresponds to the horizontal solution we already have.

So we are interested in p = 24/5
q = 3 - 24/5 = -9/5 (must be negative)

slope of tangent = -p/(4q) = 2/3
equation of tangent is:
y - 3 = 2/3 * (x - 12)
3y = 2x - 15

*NOTE TO ERIK BELOW*
Draw a sketch of the ellipse and the point (12,3). You'll see that the tangent can NEVER have a negative slope.

Answer 2

Find the y intercepts:

Substituting 0 for x in the equation leaves the quadratic 4y^2 = 36, which gives y intercepts of +/- 3. Geometrically, since the line intercepts (12, 3), this leaves a horizontal tangent line that runs through the top of the ellipse with the equation y = 3.

At the bottom of the ellipse, the line passes from (0, -3) through (12, 3). Looking at the graph, it is easy to deduce that the slope intercept equation is y = 1/4x + (-3). Now you have the two equations!

Answer 3

You're only giving one point to pass through, so there is only one tangent line.

Okay differentiate.

2x + (8y)(y') = 0

Subtract 2x

(8y)(y') = -2x

Divide by 8y

y' = -x/4y

So then x = 12 and y = 3

y' = -12/12

y' = -1

Since you are asking for the equation I'll go on further.

Slope = -1

y = -x + b

Plug in a point you have, (12,3)

3 = -12 + b

b = 15

y = -x + 15 is your tangent line

I've already given you the derivative so you can find the slope of every single point you need.

The equation is the only thing you need to find if you need more equations.

THE BELOW ARE EXTRA DETAILS YOU MIGHT WANT!

As extra detail, When I tried differentiating, we were looking for dy/dx so I used implicit differentiate to make it easier.

x^2 = (x)^2

Using chain rule you get 2x

4y^2 = (2y)^2

Using chain rule you get 2(2y)(dy/dx)

Thus (4y)(y')

Answer 4

Think of y as a function of x and:

x^2 + 4*y(x^2) = 36

differentiate:

d/dx [x^2 + 4*y(x^2) = 2x + 8y * dy/dx = 0

from this:

dy/dx = -2x / 8y = -x/4y

which implies that dy/dx = -12/12 = -1 at point (12,3) so the equation of that line would be:

y - 3 = -1 * (x - 12) or

y = 15 - x

Answer 5

Can't there be only one tangent line per point per function? It's the nature of a tangent? No?

Answer 6

its y=3 and y=.7x-5 aproxamatly