parametrics :) (1 Viewer)

jnney · Oct 2, 2011

Points P(10p, 5p^2) and Q (10q, 5q^2) lie on the parabola x^2=20y. Find the equation of the locus of the midpoint of PQ if pq=-2

I got 10y+20=x^2

the answer is 10y-100=x^2

>:[

SpiralFlex · Oct 2, 2011

jnney said:
Points P(10p, 5p^2) and Q (10q, 5q^2) lie on the parabola x^2=20y. Find the equation of the locus of the midpoint of PQ if pq=-2

I got 10y+20=x^2

the answer is 10y-100=x^2

>:[

Who on Earth answers maths questions at night time?

Let's find the midpoint first,

$M(x,y)=\frac{10p+10q}{2}, \frac{5p^2+5q^2}{2}$

$x=5(p+q).....(1)$

$y=\frac{5}{2}(p^2+q^2).....(2)$

Using the identity,

$a^2+b^2=(a+b)^2-2ab$

From 2,

$y=\frac{5}{2}[(p+q)^2-2pq]$

From 1,

$\frac{x}{5}=p+q$

Into 2,

$y=\frac{5}{2}[(\frac{x}{5})^2+4]$

$y=\frac{5}{2}*\frac{x^2}{25}+10$

$y=\frac{x^2}{10}+10$

$10y-100=x^2$

EDIT: Technically this question made me lose 1 hour. (Daylight savings)

s2 SEductive · Oct 2, 2011

^Beautiful =]

SpiralFlex · Oct 2, 2011

s2 SEductive said:
^Beautiful =]

Do not use such strong emotions human.

interesting · Oct 2, 2011

why can't we just substitute 1 into 2 instead of introducing pq?

interesting · Oct 2, 2011

nevermind, understand it now!

parametrics :) (1 Viewer)

jnney

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SpiralFlex

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s2 SEductive

MostDefinitelyIncredible

SpiralFlex

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interesting

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