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# Coordinate Systems Help? Hyperbolae watch

1. Attachment 616408Can someone help me with this question? I managed to do part i where i got
y=xp^2 -cp^3 +c/p
For the second part I equated the y from the original equation to the y in the equation for normal and now I'm stuck in big old mess.
2. Hello? Anyone there ?
Hello? Anyone there ?
The attachment isn't working for me.
Attachment 616408Can someone help me with this question? I managed to do part i where i got
y=xp^2 -cp^3 +c/p
For the second part I equated the y from the original equation to the y in the equation for normal and now I'm stuck in big old mess.
im assuming your "big old mess" is correct, did you get as far as solving a quadratic using the quadratic formula?
5. (Original post by B_9710)
The attachment isn't working for me.
6. (Original post by DylanJ42)
im assuming your "big old mess" is correct, did you get as far as solving a quadratic using the quadratic formula?
Here's what I got so far:
Here's what I got so far:
looks okay yea, write it in the form and solve for x (youll need the quadratic formula and it might be tricky)

you should be expecting a quadratic here since the line crosses the curve(s) at 2 places, and the other coordinates they want you to find
8. (Original post by DylanJ42)
looks okay yea, write it in the form and solve for x (youll need the quadratic formula and it might be tricky)

you should be expecting a quadratic here since the line crosses the curve(s) at 2 places, and the other coordinates they want you to find
I got (2cp , 3cp^3 + c/p) and (-c/p^3 , -cp^3)

Attachment 616438
Attached Images

I got (2cp , 3cp^3) and (-c/p^3 , -cp^3). Is that right? I'm not able to post my solution atm

Attachment 616438
you made a little slip on the LHS answer , otherwise thats correct

you should always notice in these type of questions (when it says to find the other coordinate) that one of the x values you get from the quadratic should be the original x value you were given, in this case is a value from the quadratic which you used originally the find the equation of the line so it can be ignored since youre only concerned with the other coordinate
10. (Original post by DylanJ42)
you made a little slip on the LHS answer , otherwise thats correct

you should always notice in these type of questions (when it says to find the other coordinate) that one of the x values you get from the quadratic should be the original x value you were given, in this case is a value from the quadratic which you used originally the find the equation of the line so it can be ignored since youre only concerned with the other coordinate

That last bit is not making much sense to me?
[/b]

That last bit is not making much sense to me?
You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
12. (Original post by B_9710)
You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
thank you, I couldnt think of how to explain that well
13. (Original post by B_9710)
You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
Which quadratic equation are you referring to? Is it new equation you get from equating the Cartesian to the tangent equation?
Which quadratic equation are you referring to? Is it new equation you get from equating the Cartesian to the tangent equation?
yea, the quadratic you solved to get the two x values, in this question one of the x values will be x=cp since you used the point P(cp, c/p) at the start to find the equation of the normal passing through P.

Its something you should be aware of as its a nice way to know whether the x values you find from the quadratic are correct.
15. (Original post by DylanJ42)
yea, the quadratic you solved to get the two x values, in this question one of the x values will be x=cp since you used the point P(cp, c/p) at the start to find the equation of the normal passing through P.

Its something you should be aware of as its a nice way to know whether the x values you find from the quadratic are correct.
Oh snap. So were the two lines intersect are the roots of the new equation. As you already know the curve already intersect/touches at x=ct you know that it should be a solution to the new equation. Thank!
So were the two lines intersect are the roots of the new equation
yes exactly, perfect

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