Paramateric equations help! (1 Viewer)

[Prazo1994]

Hi guys,
could you help me with this question:
The tangent to the point P(2ap, ap^2) on the parabola x^2=4ay cuts the x-axis in A and the y-axis on B.
i) Find the coordinates of M, the midpoint of A and B in terms of P
ii) show that the locus of M is a parabola

thanks!!!!!!!

 

[Sanjeet]

i) Differentiate, sub 2ap into x to get the gradient of the tangent (should end up being p). sub into point-gradient formula and sub x = 0 and y = to get the coordinates of B and A respectively, then use midpoint formula for these two points to get M
ii) using the x and y coordinates of M, find p in terms of x and sub that into y to get the locus

 

[cineti970128]

Hello
Well first of all you need to find the equation of the tangent.
the parabola equation can be differentiated to y' = 2x/4a
therefore when you sub x = 2ap to find the gradient of the tangent
m = p
using gradient and a point formula
(y-ap^2)/(x-2ap) = p

thus y = px - ap^2 (after expansion and simplification

then the coordinate of A (x1, 0) since y = 0 is x int

thus x1 = (0 + ap^2)/p = ap
thus coordinate of A (ap, 0)


Coordinate of B is when x = 0
therefore y = p(0) - ap^2

thereofre coordinate of B (0,-ap^2)

therefore Midpoint = ((0 + ap)/2, (-ap^2 + 0)/2 )
M (ap/2 , (-ap^2)/2 )

thus M has x coordinate moving at ap/2
and y coordnate moving at (-ap^2)/2 as parameter p changes

1. x = ap/2 ,,,,,, 2x/a = p

2. y = (-ap^2)/2

Now solving simultaneously use equation 1 into 2

y = (-a(2x/a)^2)/2
thus y = (-4x^2)/2
thus y = -2x^2

this is ofc parabola ^^
Hope this helps