Point of Intersection of two lines

[Jumpsmash]

Two tangents to an ellipse meet at a point T, find the coordinates of T.

The two equations are
(bcosΘ)x + (asinΘ)y= ab
(-bsinΘ)x + (acosΘ)y= ab


This has been really frustrating me as I feel it should be simple, but with the trigonometric parts I keep hitting a wall.

I have tried finding x in the first equation, which I worked out as X=(ab-asinΘy)/bcosΘ , and then substituting into the second equation to get a value for y, but I couldn't get it to condense to a reasonable form. I don't think this is the right approach and would appreciate someone pointing me in the right direction of how I should approach it.

 

[eumyang]

I have tried finding x in the first equation, which I worked out as X=(ab-asinΘy)/bcosΘ , and then substituting into the second equation to get a value for y, but I couldn't get it to condense to a reasonable form.

After substituting into the 2nd equation, multiply both sides by cosΘ. You'll be able to use the Pythagorean identity sin² Θ + cos² Θ = 1 to simplify things quite a bit.

 

[Jumpsmash]

If I work it through I seem to get once x has been substituted into the second equation

aby(cos^2Θ+sin^2Θ)= ab^2(cosΘ+sinΘ)
y=b(cosΘ+sinΘ)

I've quite probably lost the plot, but I have repeated this several times and come up with this. And given this result multiplying by cosΘ doesn't seem to help me. When I substitute this into the original equation I get an even more unwieldly result, and considering I have to use this point subsequently in what I'm doing I don't think it's right.

 

[HallsofIvy]

What you give for x simply isn't correct.
Multiply the first equation, (bcosΘ)x + (asinΘ)y= ab, by cosΘ, multiply the second equation, (-bsinΘ)x + (acosΘ)y= ab, by -sinΘ and add the two equations to eliminate y. There will be NO "b" in the result for x.

 

[eumyang]

What you give for x simply isn't correct.

If you're referring to the OP:
X=(ab-asinΘy)/bcosΘ
... well, I got this too.

If I work it through I seem to get once x has been substituted into the second equation

aby(cos^2Θ+sin^2Θ)= ab^2(cosΘ+sinΘ)
y=b(cosΘ+sinΘ)

I've quite probably lost the plot, but I have repeated this several times and come up with this. And given this result multiplying by cosΘ doesn't seem to help me. When I substitute this into the original equation I get an even more unwieldly result, and considering I have to use this point subsequently in what I'm doing I don't think it's right.

What Halls is suggesting (linear combinations) is probably easier. But if you want to stick with substitution:
$$x = \frac{ab - a \sin\theta y}{b \cos \theta}$$

Substitute this into the second equation:
$$-b\sin\theta \left( \frac{ab - a \sin\theta y}{b \cos \theta}\right) + a\cos\theta y = ab$$

Cancel out the b:
$$-\sin\theta \left( \frac{ab - a \sin\theta y}{\cos \theta}\right) + a\cos\theta y = ab$$

This is where I'm suggesting to multiply by cosΘ:
$$-\sin\theta \left( ab - a \sin\theta y \right) + a\cos^2 \theta y = ab\cos\theta$$

From here you should be able to solve for y. It looks like you did, because you got
y = b cos Θ + b sin Θ
which is what I got.

Now substitute into
$$x = \frac{ab - a \sin\theta y}{b \cos \theta}$$
and get
$$x = \frac{ab - a \sin\theta (b\cos\theta + b\sin\theta)}{b \cos \theta}$$
and simplify.

 

[Jumpsmash]

Many thanks to both of you. I see the method by Halls and got the solutions through that. It did indeed seem quicker. With regard to the substitution, my apologies I did multiply through by cosΘ without realising and in fact reached the stage you showed. It was from there that I could not simplify it and thought I must have the wrong solution. From using the other method it appears x=a(cosΘ-sinΘ); but I still don't see how to simplify the equation for x from the subsitution to this form.

 

[eumyang]

You mean simplifying x from here?
$$x = \frac{ab - a \sin\theta (b\cos\theta + b\sin\theta)}{b \cos \theta}$$

Distribute:
$$x = \frac{ab - ab \sin\theta \cos\theta - ab\sin^2\theta}{b \cos \theta}$$

Factor out an ab and cancel out the b:
$$x = \frac{a(1 - \sin\theta \cos\theta - \sin^2\theta)}{\cos \theta}$$

Use the Pythagorean identity again (sin² θ + cos² θ = 1 -> 1 - sin² θ = cos² θ):
$$x = \frac{a(\cos^2 \theta - \sin\theta \cos\theta)}{\cos \theta}$$

Can you finish from here?

 

[Jumpsmash]

Apologies for delay thought I sent this. Yes I can, many thanks.