What is the Equation of the Tangent Line on an Ellipse?

[Physicsissuef]

Homework Statement

The equation of tangent is given t:2x+3y-2=0 and the equation of elipse E:x^2+4y^2=K
Find "a" and "b" and the coordinates of touching point D.

Homework Equations

equation of elipse: $$b^2x^2+a^2y^2=a^2b^2$$

equation for touching: $$a^2k^2+b^2=n^2$$

equation for K (if it is circular) : $$x^2+y^2=a^2$$

The Attempt at a Solution

Actually, I don't know what is K? Is it the equation of circular? Thanks.

 

[HallsofIvy]

That question is very poorly phrased- I don't know if it was the original wording of the problem or your copy: It doesn't make sense to say "The equation of tangent is" without saying tangent to what! Since we are given the equation of an ellipse next, i assume that line is tangent to the given ellipse but if so, it would have been good to say that! I also dislike saying "find a and b" when a and b hadn't been mentioned yet.

I presume the problem is really something like "The equation of an ellipse is x²+4y²=K and a tangent to that ellipse is given by 2x+3y-2=0. What are the lengths of the semi-major and minor axes (the "a" and "b" in the standard equation of an ellipse b²x²+ a²y²= a²b²) and what are the coordinates of the point where the tangent line touches the ellipse."

As for "I don't know what is K?", that's the whole question! If you were given the entire equation of the ellipse, you would just have to put it in standard form to find a and b and that would not involve the tangent line. You have to find K so that the line and ellipse are tangent.

There is, of course, no reason to assume that the figure is a circle, especially after they have told you specifically that it is an "ellipse".

I have no idea what your "equation of touching" is because you haven't said what "k" (is that the same as "K") or "n" mean. I do know that, at any point (x,y) on an ellipse, the tangent line has slope y'= -b²x/(a²y). What is the slope of the given tangent line? What points on the ellipse have a tangent line with that slope (that will depend on K)? At what points does the given line cross the ellipse (that will also depend on K)? That gives you two equations you can solve for K. After that, it should be easy to complete the problem.

 

[Physicsissuef]

Sorry, but I don't know the names of semi-major and minor axes. Yes, all you said is true. I need to find the coordinates of the point where the tangent line touches ellipse. Also the tangent "t" is tangent on the ellipse... What should I substitute for K? I don't thinks so that K=k, in my task that is not mentioned.

 

[tiny-tim]

Hi Physicsissuef!

x² + 4y² = K means that a² = K, b² = K/4 (because x²/K + 4y²/K = 1).

(And a and b are the semi-major and semi-minor axes …the other way round, of course, if a < b.)

 

[Physicsissuef]

and how I will find K?

 

[HallsofIvy]

By solving the problem!

What is the slope of the given tangent line? Where, on the ellipse is dy/dx equal to that?

 

[Physicsissuef]

I hear first time for slope... Is there another way?

 

[Pythagorean]

I hear first time for slope... Is there another way?

slope is the derivative, like dy/dx. This probably isn't a prereq for your problem though.

in terms of algebra, slope is also m in:
y = mx+b, the standard equation of a sraight line.

 

[tiny-tim]

Hi Physicsissuef!

"slope" just means "tangent" (of the angle relative to the x-axis).

I think HallsofIvy is trying to avoid saying "the tangent of the tangent" … !

 

[Physicsissuef]

Can somebody start, please? I still can't imagine what is slope or what is the equation of slope...

 

[Pythagorean]

Can somebody start, please? I still can't imagine what is slope or what is the equation of slope...

it can be (y2-y1)/(x2-x1)

where (x1,y1) and (x2,y2) are two points on the line.

 

[Physicsissuef]

Aaa, I understand... But is K=k ? How can I use k here?

 

[Physicsissuef]

$$K^2*(\frac{-2}{3})^2+\frac{K}{4}=(\frac{2}{3})^2$$

Actually for K I get $$K^2=\frac{64}{73}$$ and in my book K=100.
Help please.

 

[tiny-tim]

Hi Physicsissuef!

What is dy/dx for the line?

What is dy/dx at a typical point (x,y) on the ellipse?

Then … ?

 

[Physicsissuef]

For first time I hear about dy/dx... Did I solve the problem right?

 

[HallsofIvy]

Then what in the world course are you taking where you were given this problem?

If you don't know about slope and you don't know about dy/dx, what do you know about "tangent lines"?

 

[tiny-tim]

$$K^2*(\frac{-2}{3})^2+\frac{K}{4}=(\frac{2}{3})^2$$

Actually for K I get $$K^2=\frac{64}{73}$$ and in my book K=100.
Help please.

Hi Physicsissuef!

Can you show us how you got that first equation … difficult to help without seeing where you went wrong.
​

If you haven't done dy/dx, then I suggest a substitution:

Put Y = 2y.

Then x² + 4y² = K becomes x² + Y² = K, and 2x + 3y - 2 = 0 becomes 2x + 3Y/2 - 2 = 0.

So you now have a circle instead of an ellipse, and it's fairly easy to see which radius is perpendicular to the line!

So you solve the problem in x and Y, and then convert back to x and y.

Alternatively … completely different method … find the points where the line crosses the ellipse, which gives you a quadratic equation which has either 0 1 or 2 real solutions. You want the line to be a tangent, so you want exactly one real solution.

What is the condition on K for there to be exactly one solution to that quadratic equation?

 

[drpizza]

I don't think he's looking for a solution involving dy/dx :) (Assuming he meant to post in the pre-calculus forum!)

physicsissuef, 100 doesn't sound right for "K". Assuming you're referring to the K that would make the equation of the ellipse x²+4y²=100. The line, 2x + 3y - 2 = 0 would go right through that ellipse; not touch it tangentially.

Unfortunately, I don't have a clue what your "equation for touching" refers to. Also, not to be too picky, but "y=mx+b" is called the slope-intercept form of the equation of a line (not the standard form.) If you rearrange the equation of your line into this form, you should be able to find the slope of the line. (m stands for slope, and b stands for y-intercept.) If you use that, plus HallsofIvy's first post in this thread, you should be able to solve it.

 

[Physicsissuef]

I know what is slope from the post of Pythagorean. (It was just translation problem).
dy/dx is also maybe some mistranslation but I don't know what is actually d ?
I solved the problem like this (if K is just variable):

$$2x+3y-2=0$$

$$3y=-2x+2$$

$$y=\frac{-2}{3}x+\frac{2}{3}$$

Out of here $$k=\frac{-2}{3}$$ and $$n=\frac{2}{3}$$

Now:

$$x^2+4y^2=K$$

$$\frac{x^2}{K}+\frac{y^2}{\frac{K}{4}}=1$$

$$a^2=K ; b^2=\frac{K}{4}$$

Now in the equation (which is condition for touching of some line to the ellipse, in this case it is the tangent given with t:2x+3y-2=0).

$$a^2k^2+b^2=n^2$$

I substitute for all values:

$$K^2*(\frac{-2}{3})^2+\frac{K}{4}=(\frac{2}{3})^2$$

And I get $$K^2=\frac{64}{73}$$ which is not same with the result of my textbook.

The result of my textbook is (the touching point) D(8,3) and a=8, b=3

 

[tiny-tim]

Now:

$$x^2+4y^2=K$$

$$\frac{x^2}{K}+\frac{y^2}{\frac{K}{4}}=1$$

$$a=K ; b=\frac{K}{4}$$

Hi Physicsissuef!

Nooo … b = K/2.

 

[Physicsissuef]

it doesn't change anything about the result.

$$a=\sqrt{K}$$, $$b=\sqrt{\frac{K}{4}$$

 

[tiny-tim]

Well … I get K = 4/5.

 

[Physicsissuef]

Ohh. I have mistake.
It should be:
$$K*(\frac{-2}{3})^2+\frac{K}{4}=(\frac{2}{3})^2$$
Since $$a^2=K$$

then the result is $$K=\frac{16}{25}$$
$$a_1=\frac{4}{5} ; b_1=\frac{2}{5}$$ ; $$a_2=\frac{-4}{5} ; b_2=\frac{-2}{5}$$

 

[tiny-tim]

The result of my textbook is (the touching point) D(8,3) and a=8, b=3

Physicsissuef … it's 2x + 3y - 25 = 0.

 

[Physicsissuef]

Yes, also the result of my book for a and b is a=10, b=5
Probably they forgot to write 5 next to 2 so it is 25.

$$2x+3y-25=0$$

$$3y=-2x+25$$

$$y=\frac{-2}{3}x+\frac{25}{3}$$

so

$$K*(\frac{-2}{3})^2+\frac{K}{4}=(\frac{25}{3})^2$$

$$K=100$$

$$a=\pm 10 ; b=\pm 5$$

Thanks for all guys...