# Tangent lines to an ellipse.

1. Feb 26, 2008

### Rossinole

Find the tangent lines to the ellipse x^2 + 7y^2 = 8 at the point (3,0)

Slope-intercept form: y=mx+b

I know you have to differentiate the equation implicitly to get the slope, but you come across a zero in the denominator and that has me stumped.

2. Feb 26, 2008

### sutupidmath

Show us what u did? I mean show all of your work, and point out where you're stumpt, so someone will point you on the right direction!!

3. Feb 26, 2008

### sutupidmath

Welcome to pf, by the way! It is a pf's policy not to give answers, but to give hints, so since you are on the homework forum you are supposed to do your homework on your own, while people here will only give hints to you!

4. Feb 26, 2008

### D H

Staff Emeritus
What kind of line can't be described in this form?

5. Feb 26, 2008

### Rossinole

x^2+7y^2=8 (3,0)

The derivative is 2x + 14y(dy/dx)=0. To get dy/dx by itself, first subtract 2x on both sides and you get 14y(dy/dx) = -2x. Then divide by 14y on both sides and you get (dy/dx)=-2x/14y. Plugging in the points (3,0) into the equation gives you 6/0, which is undefined.

This is where I'm stuck. I know there has to be another way to approach this, I just need someone to point me in the right direction.

6. Feb 26, 2008

### D H

Staff Emeritus

7. Feb 26, 2008

### sutupidmath

for what type of line we say has no slope. Or roughly speaking has a slope of "infinity"?

8. Feb 26, 2008

### Rossinole

When you solve for dx/dy, you get 2x(dx/dy)+14y=0, 2x(dx/dy)=-14y, (dx/dy)=-14y/2x, (dx/dy)=-14(0)/2(3), (dx/dy)=0/6, (dx/dy)=0. So you end with the tangent line being y=0. There's still supposed to be at least one more line. This, again, is what is confusing me. How would I go about finding more than one tangent line?

9. Feb 26, 2008

### D H

Staff Emeritus
There is only one tangent line to an ellipse at any point on the ellipse. An ellipse is a simple curve (it doesn't cross itself; the symbol for infinity or the digit "8" are examples of curve that cross themselves) and it doesn't have a cusp.

10. Feb 26, 2008

### Rossinole

I'm pretty sure the point isn't on the ellipse. I'm sorry I guess I should have made that more clear.

11. Feb 26, 2008

### sutupidmath

And you are letting us know just now????
well then the eq of the line is y=mx+b, let this line go throught the point (3,0) so we get

0=3m+b, now the slope of the line at any point in the tangent line as you computed is (dy/dx)=-2x/14y so
0=3(-2x/14y)+b, now this is an eq of the line that goes through (3,0) try to find at what points does this line touch the elipse!

Last edited: Feb 26, 2008
12. Feb 26, 2008

### D H

Staff Emeritus
Doh! We should have seen that. Yes, there are two such lines, neither one is vertical.

13. Feb 26, 2008

### Rossinole

Shouldn't it be the (dx/dy) in the slope, not (dy/dx)? But my guess for that would be plug in the points into the equation and you get 0=3(6/0)+b, which, you can't do..If you did it (dx/dy), you'd end up with b=0?

14. Feb 26, 2008

### D H

Staff Emeritus
It might be easier to work with the point-slope form for a line:

$$\frac {y-y_1}{x-x_1} = m$$

where the reference point $(x_1,y_1)[/tex] is some point on the ellipse. You already have an expression for [itex]m$:

$$\frac{dy}{dx}=-\,\frac{\phantom{1}2x_1}{14y_1}$$

You also know the coordinates of another point on the line: $(3,0)$.

See where you can go with this.

15. Feb 26, 2008

### Rossinole

Haha, I really don't know where to go with it. Substituting the coordinates of the other point on the line and setting it equal to (dy/dx) is my first guess. 0-y1/3-x1=-(2x/14y). But even then I don't know where to go with that.

Is this in anyway similar to solving for two tangent lines of a parabola?

16. Feb 26, 2008

### D H

Staff Emeritus
First, put the expression for the known slope of the line into the equation of the line. Using the point-slope form

$$\frac {y-y_1}{x-x_1} = -\,\frac{x_1}{7 y_1}$$

(I simplified 2/14 to 1/7 in the above). You know an (x,y) pair, (3,0). Put this into the equation. You will be left with a relation between $x_1$ and $y_1$. You already have another relation between these parameters: the equation of the ellipse.

17. Jan 4, 2011