Hyperbola chord bisected at a given point

[DarkLord76865]

The problem goes like this:
there is a point [math]T (5, 1)[/math] which bisects chord of hyperbola [math]4x^2-9y^2=36[/math]I need to find the equation of line of that chord.

I've found this on the Internet:

It gives correct solution [math]20x-9y-91=0[/math] but I don't understand why. What is T and S1? T looks like an equation for tangent on a hyperbola in given point, and S1 looks like an equation for hyperbola. None of this makes sense to me because given point [math]T (5, 1)[/math] is NOT on given hyperbola.

I've also found this video, but I have trouble understanding anything the guy speaks.

Hope someone can clarify

Thanks

 

[topsquark]

An English speaker listening to someone with a heavy Indian accent is an art form. I went to Alfred University for Ceramics for two years and I went to a Ceramics class just after taking Spanish. It was quite an experience. Still you never quite forget.

Anyway. He was working on a parabola and you have a hyperbola. But the method is similar.

This is a guide, not a solution. More or less I've given you all the steps but you need to fill in the work between some of them. Let us know how it goes.
[imath]4x^2 - 9y^2 = 36[/imath]

So
[imath]y^2 = \dfrac{4x^2 - 36}{9}[/imath]

Pick two points on the hyperbola and call them [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. These will be the points on the hyperbola that (5, 1) is the midpoint for.

Then we know
[imath]y_1^2 = \dfrac{4x_1^2 - 36}{9}[/imath]

and
[imath]y_2^2 = \dfrac{4x_2^2 - 36}{9}[/imath]


Subtract these to get
[imath]y_1^2 - y_2^2 = \dfrac{4}{9}(x_1^2 - x_2^2)[/imath]

This becomes
[imath]\dfrac{y_1 - y_2}{x_1 - x_2} = \dfrac{4}{9} \cfrac{x_1 + x_2}{y_1 + y_2}[/imath]

Now. The line y = mx + b going through (5, 1) also goes through [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. Choosing the points [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath], what is the slope m of the line?

So. (5, 1) is the midpoint between [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. So
[imath]5 = \dfrac{x_1 + x_2}{2}[/imath]

and
[imath]1 = \dfrac{y_1 + y_2}{2}[/imath]

Divide 5/1. What do you get?

Compare your formulas for m and 5/1 with the equation [imath]\dfrac{y_1 - y_2}{x_1 - x_2} = \dfrac{4}{9} \cfrac{x_1 + x_2}{y_1 + y_2}[/imath]. What can you do to simplify this equation?

Okay, so if you've got that you have a value for m. Use the fact that y = mx + b and that the point (5, 1) is on this line to get a value for b.

Let us know if you run into any problems with this. Show us what you've got and we can work with you.

-Dan

 

[Otis]

None of this makes sense to me because given point T(5,1) is NOT on given hyperbola.

Hi DarkLord. The chord is a line segment. Its endpoints lie on the hyperbola. The given point lies on the chord, midway between the endpoints. The right branch of the hyperbola is plotted below, with the chord and given midpoint.


[imath]\;[/imath]

 

[DarkLord76865]

An English speaker listening to someone with a heavy Indian accent is an art form. I went to Alfred University for Ceramics for two years and I went to a Ceramics class just after taking Spanish. It was quite an experience. Still you never quite forget.

Anyway. He was working on a parabola and you have a hyperbola. But the method is similar.

This is a guide, not a solution. More or less I've given you all the steps but you need to fill in the work between some of them. Let us know how it goes.
[imath]4x^2 - 9y^2 = 36[/imath]

So
[imath]y^2 = \dfrac{4x^2 - 36}{9}[/imath]

Pick two points on the hyperbola and call them [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. These will be the points on the hyperbola that (5, 1) is the midpoint for.

Then we know
[imath]y_1^2 = \dfrac{4x_1^2 - 36}{9}[/imath]

and
[imath]y_2^2 = \dfrac{4x_2^2 - 36}{9}[/imath]


Subtract these to get
[imath]y_1^2 - y_2^2 = \dfrac{4}{9}(x_1^2 - x_2^2)[/imath]

This becomes
[imath]\dfrac{y_1 - y_2}{x_1 - x_2} = \dfrac{4}{9} \cfrac{x_1 + x_2}{y_1 + y_2}[/imath]

Now. The line y = mx + b going through (5, 1) also goes through [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. Choosing the points [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath], what is the slope m of the line?

So. (5, 1) is the midpoint between [imath](x_1, y_1)[/imath] and [imath](x_2, y_2)[/imath]. So
[imath]5 = \dfrac{x_1 + x_2}{2}[/imath]

and
[imath]1 = \dfrac{y_1 + y_2}{2}[/imath]

Divide 5/1. What do you get?

Compare your formulas for m and 5/1 with the equation [imath]\dfrac{y_1 - y_2}{x_1 - x_2} = \dfrac{4}{9} \cfrac{x_1 + x_2}{y_1 + y_2}[/imath]. What can you do to simplify this equation?

Okay, so if you've got that you have a value for m. Use the fact that y = mx + b and that the point (5, 1) is on this line to get a value for b.

Let us know if you run into any problems with this. Show us what you've got and we can work with you.

-Dan

Sorry, I've never answered. My friend showed me some other solution so I kinda forgot about it until I was going through old email. Yes, I perfectly understand your solution, it just makes sense and is much easier than what my friend showed me. It is actually so easy I can't believe I'm too stupid I didn't manage to solve it myself. Thanks