conics question (1 Viewer)

[bboyelement]

show that the line 2y+x+5 = 0 is a tangent to the ellipse (x^2)/9 + (y^2)/4 = 1 . find the coordinates of the point of contact.

thanks

 

[STx]

First, multiply the eq'n of the ellipse by 36 to get rid of both those denominators and get something more user-friendly,

4x^2+9y^2 = 36 (1)

then substitute x=-5-y (from the eq'n for the line) into (1)

You should get a quadratic in terms of y.

You then need to show that the discriminant of this quadratic is zero, which means that the quadratic has a double root. That means that there is only one point that satisfies both the eq'n for the line and the ellipse, i.e. that the line is a tangent to the ellipse.

Finding the root of the quadratic gives you the y-coordinate of the point of contact. Substitute it back into the eq'n of the line to find the x-coord.

yep, only thing is sub substitute x=-5-2y (from the eq'n for the line) into (1)

 

[bboyelement]

thanks so much guys i was wondering if there was any other way of doing it

 

[hyparzero]

Heres my way:

Tangent Eqn: 2y + x + 5 = 0
Rearrange 2y = - 5 - x
y = - (5 + x)/2
Therefore (dy/dx) = - 1/2 (1)

Ellipse Eqn: x²/9 + y²/4 = 1
Rearrange: 4x² + 9y² = 36
Differentiate all sides: d(4x² + 9y²) = d(36)
=> 8x + 18(dy/dx) = 0
Rearrange: (dy/dx) = - 8x/18
= - 4x/9 (2)

Since the tangent touches the curve at P(x,y)
We have (1) = (2)

=> - 1/2 = - 4x/9
1/2 = 4x/9
Rearrange: 8x = 9
Therefore, the x-coordinate of the Point of Contact is (8/9)

Finding the y-coordinate should be easy, sub x= 8/9 into 2y + x + 5 = 0

Hope I've helped!

 

[_ShiFTy_]

Heres my way:

Tangent Eqn: 2y + x + 5 = 0
Rearrange 2y = - 5 - x
y = - (5 + x)/2
Therefore (dy/dx) = - 1/2 (1)

Ellipse Eqn: x²/9 + y²/4 = 1
Rearrange: 4x² + 9y² = 36
Differentiate all sides: d(4x² + 9y²) = d(36)
=> 8x + 18(dy/dx) = 0
Rearrange: (dy/dx) = - 8x/18
= - 4x/9 (2)

Since the tangent touches the curve at P(x,y)
We have (1) = (2)

=> - 1/2 = - 4x/9
1/2 = 4x/9
Rearrange: 8x = 9
Therefore, the x-coordinate of the Point of Contact is (8/9)

Finding the y-coordinate should be easy, sub x= 8/9 into 2y + x + 5 = 0

Hope I've helped!

Someone please correct me if im wrong, but i dont think this method really proves that 2y + x + 5 = 0 is the tanget to that ellipse, because the ellipse can have 2 tangents for any given gradient

 

[pLuvia]

Yeh but for any tangent from the ellipse there is only one gradient

So hyparzero's way is valid

because the ellipse can have 2 tangents for any given gradient

I think you mean the ellipse can have 2 tangents from an external point

This doesn't make sense either, lets say you get the dy/dx of the ellipse, if you sub in an x value you will only get one gradient

 

[_ShiFTy_]

Yeh but for any tangent from the ellipse there is only one gradient

So hyparzero's way is valid



I think you mean the ellipse can have 2 tangents from an external point

This doesn't make sense either, lets say you get the dy/dx of the ellipse, if you sub in an x value you will only get one gradient

But say you have the basic ellipse x^2/a^2 + y^2/b^2 = 1
Then wouldnt the tangents at both y = ±b have the same gradient?

Because when u differentiate the ellipse from the example in the first post, dy/dx = -8x/18y ... so if the gradient of the line is -1/2, then 8x/18y = 1/2, 16x = 18y ... so there would be 2 points with tangents that satisfy that equation

 

[pLuvia]

But aren't you saying that for each gradient there is two tangents that can arise from it?

so there would be 2 points with tangents that satisfy that equation

But here you're saying that for an ellipse, from an external point there can be two tangents which is correct

Then wouldnt the tangents at both y = ±b have the same gradient?

How could they have the same gradient though if they are intersecting at an external point. In that y=±b they would just be parallel lines which don't intersect at an external point

Do you understand what I mean?

 

[_ShiFTy_]

Ok, first of all i am not referring to any external point, i am talking about the points on the ellipse

It is clear that the tangets at both y = ± b on the general ellipse (the max and min y values of the ellipse) have the same gradient which is 0 (a horizontal line)

This is the same for any other 2 corresponding points. e.g. If you are given that the gradient of a line is m=2, and this line is a tangent to a particular ellipse, you will get 2 equations


Lastly, in hyparzero's method, there is an incorrect step when he differentiated implicitly..

Differentiate all sides: d(4x² + 9y²) = d(36)
=> 8x + 18(dy/dx) = 0

There is a missing y in 18(dy/dx)..that is how he only got one solution


So the only way to prove it using hyparzero's method, is that you need to find the point of intersection (of the ellipse and the line) and then sub it into 16x=18y (from my previous post)

 

[pLuvia]

ok ok now I know what your saying

 

[c0okies]

is there a faster way of finding the coordinates instead of subbing the eqn in?

 

[Riviet]

is there a faster way of finding the coordinates instead of subbing the eqn in?

There are other ways of getting the co-ordinates like changing the ellipse into parametric form, but that would take more time than just subbing the equation into the ellipse.
I'm personally finding this conics topic very tedious, especially with all the algebra; you really have to be solid with your algebra, one silly mistake and that'll stuff up the rest of your answer.

 

[c0okies]

yea; its simple but a pain in the neck, because the workingout is soo long

 

[bboyelement]

I'm personally finding this conics topic very tedious, especially with all the algebra; you really have to be solid with your algebra, one silly mistake and that'll stuff up the rest of your answer.

true and all the trig identities ... sometimes it would take ages just to get one question done ... if you dont know you trig very well ... some of those questions in the fitzpatrick are killing me