The curve 12x² - 5y² = 7 intersects the line 2p²x - 5y = 7 at the point (1,p).

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pancakes_fn

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The curve 12x² - 5y² = 7 intersects the line 2p²x - 5y = 7 at the point (1,p).
(i) Find the value of p
(ii) Find the coordinates of the other point of intersection.

Here’s what I’ve got so far:
Equate both equations:
12x² - 5y² = 2p²x - 5y
Substitute (1,p):
12(1)² - 5(p)² = 2p²(1) - 5(p)
12 - 5p² = 2p² - 5p
12 - 5p² - 2p² + 5p = 0
-7p² + 5p + 12 = 0
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
p = 12/7 or p = -1
How do I continue?
How do I know which value is the correct one?
 
pancakes_fn said:
The curve 12x² - 5y² = 7 intersects the line 2p²x - 5y = 7 at the point (1,p).
(i) Find the value of p
(ii) Find the coordinates of the other point of intersection.

Here’s what I’ve got so far:
Equate both equations:
12x² - 5y² = 2p²x - 5y
Substitute (1,p):
12(1)² - 5(p)² = 2p²(1) - 5(p)
12 - 5p² = 2p² - 5p
12 - 5p² - 2p² + 5p = 0
-7p² + 5p + 12 = 0
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
p = 12/7 or p = -1
How do I continue?
How do I know which value is the correct one?
Click to expand...
You have solved the quadratic equation involving 'p' correctly.

To justify "each" value of 'p' - you need to show that (1,p) satisfies both the equations (the point of intersection) .
When p = -1, the point of intersection (1,p) becomes (1,-1)

check first equation 12*x^2 - 5*y^2 = 12 - 5 = 7 ............checks
check first equation 2*p^2 * x - 5*y = 2 + 5 = 7 ............checks

So (1,-1) belongs to both the curves and is a point of intersection.

Do the same for p = 12/7

According to your problem, what is the next thing you need to do?
 
hi, thanks for replying.
after susbstituting p=12/7 into both equations, i realized why p=12/7 is rejected for (i).
i managed to solve the (ii) and got (-1 1/2 , -2). thanks so much for your help.
 
pancakes_fn said:
The curve 12x² - 5y² = 7 intersects the line 2p²x - 5y = 7 at the point (1,p).
(i) Find the value of p
(ii) Find the coordinates of the other point of intersection.

Here’s what I’ve got so far:
Equate both equations:
12x² - 5y² = 2p²x - 5y
Substitute (1,p):
12(1)² - 5(p)² = 2p²(1) - 5(p)
12 - 5p² = 2p² - 5p
12 - 5p² - 2p² + 5p = 0
-7p² + 5p + 12 = 0
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
p = 12/7 or p = -1
How do I continue?
How do I know which value is the correct one?
Click to expand...
What you did is fine but in my opinion much too hard. You are told that (1,p) is on the line 2p²x - 5y = 7. So 2p²(1) - 5p = 7 or 2p² - 5p - 7 = 0. This yields (2p-7)(2p-1)=0. So p=7/2 or p=-1. This means that (1,p) is on the line 2p²x - 5y = 7 only when p = 7/2 and p=-1. Now verify which of these p values, if any, satisfy the the 2nd line for (1,p).
 
Its been 5 years since this post and i need the solution for part ii. 😭 pls helpp my hw is due tmrr
 
Lilypopp said:
Its been 5 years since this post and i need the solution for part ii. 😭 pls helpp my hw is due tmrr
Click to expand...
In that case, please do as we ask everyone:

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We don't just give solutions. We help you get over wherever you are stuck, so we need to see where that is.

(Homework is meant to be a chance for you to learn how to solve problems, not how to beg others for answers.)
 
Lilypopp said:
Its been 5 years since this post and i need the solution for part ii. 😭 pls helpp my hw is due tmrr
Click to expand...

You are not the original poster! So, as a new member with one post, you never posted it back on Dec. 8, 2019.
It would not make sense that you would be needing it.

You should not be responding to a post that is at least four years old, anyway.
 
lookagain said:
You are not the original poster! So, as a new member with one post, you never posted it back on Dec. 8, 2019.
It would not make sense that you would be needing it.
Click to expand...
Presumably the problem comes from a book that is still in use, and was assigned to this student; I found a more recent source with an image:

1708015968174.png

But no one "needs the solution", regardless. They need to learn how to solve such problems, not to search for a site that has the solution, in order to avoid learning.
 
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