easy parametric (1 Viewer)

[Timothy.Siu]

24c 11. from fitz
need help
P is a variable point on the parabola x²=-4y. The tangent from P cuts the parabola x²=4y at Q and R. Show that 3x²=4y is the equation of the locus of the mid-point of the chord RQ.

doesn't look hard...just cant figure out the first steps

 

[trailblazer]

24c 11. from fitz
need help
P is a variable point on the parabola x²=-4y. The tangent from P cuts the parabola x²=4y at Q and R. Show that 3x²=4y is the equation of the locus of the mid-point of the chord RQ.

doesn't look hard...just cant figure out the first steps

Woah, coincidentally im stuck on the same question.

 

[Trebla]

24c 11. from fitz
need help
P is a variable point on the parabola x²=-4y. The tangent from P cuts the parabola x²=4y at Q and R. Show that 3x²=4y is the equation of the locus of the mid-point of the chord RQ.

doesn't look hard...just cant figure out the first steps

y = - x²/4
dy/dx = - x/2
At P(- 2p, - p²) NB: a = 1 and p>0, hence both x and y coordinates are negative
dy/dx = p
Equation of tangent is: y + p² = p(x + 2p)
Since it cuts x² = 4y => y = x²/4, sub it in
x²/4 + p² = p(x + 2p)
x² + 4p² = 4px + 8p²
x² - 4px + 4p² = 8p²
(x - 2p)² = 8p²
x = 2p ± 2p√2
Midpoint of RQ gives x coordinates {(2p + 2p√2) + (2p - 2p√2)}/2 = 2p
Sub into tangent equation: (as midpoint lies on tangent)
y = p(2p) + p²
= 3p²
So our parametric equations are:
x = 2p
y = 3p²
Hence:
3x² = 12p²
= 4(3p²)
= 4y