Hard Parabola Question (1 Viewer)

[BoganBoy]

Anyone who can answer this will be worshipped by me. i am too frustrated.

The point P(4p,2p^2) and Q(4q, 2q^2) lie on a parabola. the normal at P is y=-x/p+2p^2+4 and it intersect the parabola at R. Find the Cartesian equation of the locus of the midpoint of PR.

Can some one please tell me how to get the intercept point? ive worked out that the parabola's equation is x^2=8y. thanks!!!

 

[who_loves_maths]

^ lol? first of all, someone explain to me why you needed to specify a point Q ???

 

[BoganBoy]

The question is actually longer, it is separated into like 5 parts. this one is the last part. The full question is this:

The point P(4p,2p^2) and Q(4q, 2q^2) lie on a parabola.

1. Find the Cartesian equation of the parabola

x^2=8y

2. Show that the equation of the chord PQ is (p+q)x-2y-4pq=0

3. if the chord passes through the focus of the parabola, show that the tangents at the end of the focal chord PQ are at right angle.

4. Show that the equation of the normal at P is y=-x/p+2p^2+4

5. The normal at point P intercept the parabola at Point R. Find the Cartesian Equation of the locus of the midpoint PR.

I have already done 1-4, but no idea how to do the fifth one, so please help.

 

[tez0r]

solve simultaneously with the equation of the parabola, so sub in y=x^2/8 into the equation of the normal. you'll get a quadratic, work from there, disregard the result where your normal originated

 

[who_loves_maths]

The question is actually longer, it is separated into like 5 parts. this one is the last part. The full question is this:

The point P(4p,2p^2) and Q(4q, 2q^2) lie on a parabola.

1. Find the Cartesian equation of the parabola

x^2=8y

2. Show that the equation of the chord PQ is (p+q)x-2y-4pq=0

3. if the chord passes through the focus of the parabola, show that the tangents at the end of the focal chord PQ are at right angle.

4. Show that the equation of the normal at P is y=-x/p+2p^2+4

5. The normal at point P intercept the parabola at Point R. Find the Cartesian Equation of the locus of the midpoint PR.

I have already done 1-4, but no idea how to do the fifth one, so please help.

okay... now that you've made it clear ... here's what i would have done (without taking into account of anything i might need at all from Q1-4): {ie. i might be wrong}

to find the point(s) of intersection of the normal at P and the parabola, one would normally go on to equate both equation by eliminating 'y':

ie. x^2/8 = -x/p+2p^2+4

---> x^2 + (8/p)x - 16p^2 - 32 = 0 ; which is a quadratic in 'x'.

now, you don't need to SOLVE this quadratic since that would be silly:

you're asked to find the midpoint of PR - meaning that the x-coordinate of the midpoint would simply be the sum of the roots of that quadratic equation divided by 2.

hence, sum of roots = -8/p ; ie. x-coordinate of midpoint = -4/p

since the normal is a line and the midpoint of PR lies on the line connecting P and R, then the midpoint lies on the normal. ie. substitute in the newly found x-value:

ie. y = 4/p^2 + 2p^2 + 4


Therefore, the Midpoint is: {-4/p , 4/p^2 + 2p^2 + 4}

let x = -4/p ---> p = -4/x ; subsitute this into the y-coordinate:

ie. y = 4/(16/x^2) + 2(16/x^2) + 4

-----> y = x^2/4 + 32/x^2 + 4 is the locus of the Midpoint of PR.


hope that helps and i hope that's right also... there's a distinct probability i am completely wrong on account of not taking note of anything in Q1-4...

 

[Dumsum]

you're asked to find the midpoint of PR - meaning that the x-coordinate of the midpoint would simply be the sum of the roots of that quadratic equation divided by 2.

Would you be so kind as to explain this? We were never taught that...makes life easy though...

 

[香港!]

^ it's like the x=-b\2a
to prove it's sum of roots=-b\a
divide 2 u get x=-b\2a

 

[香港!]

solve simultaneously with the equation of the parabola, so sub in y=x^2/8 into the equation of the normal. you'll get a quadratic, work from there, disregard the result where your normal originated

hihi
where did u get ur avatar?
it looks coool!

 

[who_loves_maths]

Originally Posted by Dumsum

Originally Posted by who_loves_maths
you're asked to find the midpoint of PR - meaning that the x-coordinate of the midpoint would simply be the sum of the roots of that quadratic equation divided by 2.

Would you be so kind as to explain this? We were never taught that...makes life easy though...

the x-coordinate of the Midpoint of PR is given by (x(1) + x(2))/2 ; where x(1) and x(2) are the x-coordinates of the points P and R.

but when you equate the equation of the normal with the equation of the parabola you get a quadratic, whose two solutions give you the x-coordinates of the two points of intersection between the normal and the parabola - namely the x-coordinates of P and R.

ie. x(1) and x(2) will be the two roots of the quadratic equation.

now, since all you need in this case is (x(1) + x(2))/2 , then clearly, as i said, it would be silly to actually go and solve the quadratic equation because (x(1) + x(2)) is simply the sum of the roots of that equation.

and using polynomial theory, sum of roots of quadratic = -b/a = -8/p (in this case)

therefore: x-coordinate of Midpoint is (x(1) + x(2))/2 = -4/p


hope that explains it for you Dumsum


P.S. as to whether it's taught or not in school, that is irrelevant ... problem solving involves a bit of thinking, not just regurgitation, and something as simple as sum of roots should appear intuitively obvious as it manifests itself in this case.