FP3:hyperbola Watch

Maths&physics
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.....
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what is q b asking for?
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RDKGames
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(Original post by Maths&physics)
what is q b asking for?
It's asking you to use the fact that the line is tangent to the hyperbola to derive the relation a^2m^2 = b^2 + c^2.
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(Original post by RDKGames)
It's asking you to use the fact that the line is tangent to the hyperbola to derive the relation a^2m^2 = b^2 + c^2.
It's been a while since I've done a question like this. The point they meet is where both of their gradients are the same?
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ghostwalker
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(Original post by Maths&physics)
It's been a while since I've done a question like this. The point they meet is where both of their gradients are the same?
Yes.

Though I suspect you're not thinking on the most useful track.

Since the roots of the quadratic are the x-coordinates of the points where they meet, and they only meet at one point for a tangent (for a hyperbola at least), it follows that your quadratic has only one root, hence....
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(Original post by ghostwalker)
Yes.

Though I suspect you're not thinking on the most useful track.

Since the roots of the quadratic are the x-coordinates of the points where they meet, and they only meet at one point for a tangent (for a hyperbola at least), it follows that your quadratic has only one root, hence....
how do I know there is only one x coordinate?

well, if we have an equation, if there is only 1 root - the discriminate = 0?
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RDKGames
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(Original post by Maths&physics)
how do I know there is only one x coordinate?

well, if we have an equation, if there is only 1 root - the discriminate = 0?
You sub in the eqn. of a line into the eqn. of the hyperbola.

The resultant equation entirely in x is satisfied only by the x-coordinates at which the line and the hyperbola intersect.

Since L is tangent to H, that means there is only one point of intersection (the line touches the hyperbola!) so there is only one real solution to the above equation in x.

Since this equation is a quadratic, you can immediately deduce that this implies a zero discriminant.


Standard C1 stuff; except there you dealt with a parabola instead of a hyperbola. Same reasoning though!
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(Original post by RDKGames)
You sub in the eqn. of a line into the eqn. of the hyperbola.

The resultant equation entirely in x is satisfied only by the x-coordinates at which the line and the hyperbola intersect.

Since L is tangent to H, that means there is only one point of intersection (the line touches the hyperbola!) so there is only one real solution to the above equation in x.

Since this equation is a quadratic, you can immediately deduce that this implies a zero discriminant.


Standard C1 stuff; except there you dealt with a parabola instead of a hyperbola. Same reasoning though!
"You sub in the eqn. of a line into the eqn. of the hyperbola.


The resultant equation entirely in x is satisfied only by the x-coordinates at which the line and the hyperbola intersect. "

yep, I understand that.

"Since L is tangent to H, that means there is only one point of intersection (the line touches the hyperbola!) so there is only one real solution to the above equation in x."

Ok, that makes sense because I was looking at the graph and the asymptotes and equated them with the talents but they can't be because the whole point is that the graph never meets the asymptote. therefore, there is only one point that they meet.

Since this equation is a quadratic, you can immediately deduce that this implies a zero discriminant.

perfect, thanks.
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RDKGames
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(Original post by Maths&physics)
Ok, that makes sense because I was looking at the graph and the asymptotes and equated them with the talents but they can't be
I'm not sure what you mean here. The asymptotes don't have a lot to do with anything here.

If you want to think about the question beyond its scope, then you would find that the eqn. of the line (under the condition of being tangent to the hyperbola at the point x=\alpha) gets closer and closer to either of the diagonal asymptotes as |\alpha| \to \infty.

whole point is that the graph never meets the asymptote
Well that's not entirely true. The graph y=\dfrac{x}{1+x^2} meets its own asymptote. You might want to review what an asymptote actually is.
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(Original post by RDKGames)
I'm not sure what you mean here. The asymptotes don't have a lot to do with anything here.

If you want to think about the question beyond its scope, then you would find that the eqn. of the line (under the condition of being tangent to the hyperbola at the point x=\alpha) gets closer and closer to either of the diagonal asymptotes as |\alpha| \to \infty.



Well that's not entirely true. The graph y=\dfrac{x}{1+x^2} meets its own asymptote. You might want to review what an asymptote actually is.
how does part b help with part c?
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(Original post by RDKGames)
I'm not sure what you mean here. The asymptotes don't have a lot to do with anything here.

If you want to think about the question beyond its scope, then you would find that the eqn. of the line (under the condition of being tangent to the hyperbola at the point x=\alpha) gets closer and closer to either of the diagonal asymptotes as |\alpha| \to \infty.



Well that's not entirely true. The graph y=\dfrac{x}{1+x^2} meets its own asymptote. You might want to review what an asymptote actually is.
is part a, the point where both tangents interest?
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RDKGames
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(Original post by Maths&physics)
how does part b help with part c?
H’ is a particular hyperbola whereas H is a general one.

By comparing the two, you can deduce the values of a,b.

Since we are seeking tangents, the condition of part B gives you a condition m,c must satisfy.

Since you are given a point these tangents pass through, it satisfies th line’s eqn so you can deduce another equation satisfied by m,c.

You have two eqns in two unknowns. Solve for them.
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RDKGames
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(Original post by Maths&physics)
is part a, the point where both tangents interest?
Not sure what you’re asking here.
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(Original post by RDKGames)
Not sure what you’re asking here.
This is how I understand what's going on: tan (blue lines) = tangent, and asy = asymptote. the green dot is where both tangents intersect.

part a gives as a general equation for any tangent for any hyperbola.

we know values for a and b for the question, so we can sub that into the equation, but I don't know what to do next???
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RDKGames
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(Original post by Maths&physics)
This is how I understand what's going on: tan (blue lines) = tangent, and asy = asymptote. the green dot is where both tangents intersect.

part a gives as a general equation for any tangent for any hyperbola.

we know values for a and b for the question, so we can sub that into the equation, but I don't know what to do next???
Show your working out. What have you subbed and where?

I'd rather verify you have the correct expressions at this stage.
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(Original post by RDKGames)
Show your working out. What have you subbed and where?

I'd rather verify you have the correct expressions at this stage.
so, from my understanding: b gives us an equation that we can use for part c.

after subbing a and b into it, we get: 25m^2 = 16 + c^2

the general equation for the hyperbola and tangent (this gives all and any equations for any tangents for any hyperbola, but once we sub our specific a and b values into it, we only get an equation for any and all tangents for this hyperbola): (25m^2 - 16)x^2 + 50mcx + 25(c^2 + 16) = 0
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RDKGames
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(Original post by Maths&physics)
so, from my understanding: b gives us an equation that we can use for part c.

after subbing a and b into it, we get: 25m^2 = 16 + c^2

the general equation for the hyperbola and tangent (this gives all and any equations for any tangents for any hyperbola, but once we sub our specific a and b values into it, we only get an equation for any and all tangents for this hyperbola): (25m^2 - 16)x^2 + 50mcx + 25(c^2 + 16) = 0
Ok, so 25m^2 = 16 + c^2 is one equation in two unknowns.

You don't need to touch the quadratic in x at all. That equation tells you the x-coordinates where the lines are tangent to the hyperbola -- not what we want!

You're told that tangents pass through (1,4). So the equation 4 = m + c must be satisfied as well. [you sub the coords into y=mx+c]

So solve for m,c and state the two equations.
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(Original post by RDKGames)
Ok, so 25m^2 = 16 + c^2 is one equation in two unknowns.

You don't need to touch the quadratic in x at all. That equation tells you the x-coordinates where the lines are tangent to the hyperbola -- not what we want!

You're told that tangents pass through (1,4). So the equation 4 = m + c must be satisfied as well. [you sub the coords into y=mx+c]

So solve for m,c and state the two equations.
is there any other way to do it?

and wow, that was so much simpler than I was expecting/thinking!
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(Original post by RDKGames)
Ok, so 25m^2 = 16 + c^2 is one equation in two unknowns.

You don't need to touch the quadratic in x at all. That equation tells you the x-coordinates where the lines are tangent to the hyperbola -- not what we want!

You're told that tangents pass through (1,4). So the equation 4 = m + c must be satisfied as well. [you sub the coords into y=mx+c]

So solve for m,c and state the two equations.
I'm still confused about b. "given that L is a tangent of H, show that..." how does the imply or explicitly state we should look for the discriminate?
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RDKGames
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(Original post by Maths&physics)
I'm still confused about b. "given that L is a tangent of H, so that..." how does the imply or explicitly state we should look for the discriminate?
Discriminant*

I explained this in post #7. Are you struggling to make the connection from 'this quadratic has one solution' to 'discriminant must be zero' ?
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