Parametric Equation (1 Viewer)

[lordinance]

Im so screwed up at this topic .Could someone help me please?

12)
Points p(2ap,ap^2) and Q(2aq,aq^2) lie on the parabola x^2=4ay
Show that the normal at P is given by x+py=ap^3 +2ap IN FULL STEP.

13a)
Point M(4,-8)lies on the parabola x^2 = -2y .Find the equation of the focal chord through M.


Many thanks

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[SeCKSiiMiNh]

Im so screwed up at this topic .Could someone help me please?​

12)
Points p(2ap,ap^2) and Q(2aq,aq^2) lie on the parabola x^2=4ay
Show that the normal at P is given by x+py=ap^3 +2ap IN FULL STEP.​

13a)
Point M(4,-8)lies on the parabola x^2 = -2y .Find the equation of the focal chord through M.​

Many thanks​

13a) x^2 = 4ay
differentiating both sides give:

2x = 4a dy/dx

dy/dx = x/2a

sub in x on point p

therefore dy/dx = 2ap/2a
=p

therefore, the gradient of the normal is -1/p

use your point gradient formula:

y-y1 = m (x-x1)

y- ap^2 = -1/p (x-2ap)

mutiply everything by p

so you get:

yp - ap^3 = 2ap - x

rearranging gives

x + yp = 2ap + ap^3

 

[SeCKSiiMiNh]

i forgot how to do 13b) but i think you need to find the value of "a" first, which is -1/2 since x^2 = -2y and x^2 = 4ay.

then use your formula to find the equation using your 2 points. i.e. 4, -8 and 0,-1/2

edit: the formuar being the 2 point forumla: y-y1/ x-x1 = y2 - y1/x2-x1

 

[fullonoob]

Im so screwed up at this topic .Could someone help me please?​

12)
Points p(2ap,ap^2) and Q(2aq,aq^2) lie on the parabola x^2=4ay
Show that the normal at P is given by x+py=ap^3 +2ap IN FULL STEP.​

13a)
Point M(4,-8)lies on the parabola x^2 = -2y .Find the equation of the focal chord through M.​

Many thanks​

13a) This is from what i remember >>> i've forgot this so it may not be 100%
x^2 = 4ay
since the equation is x^2 = -2y
a = -1/2
Therefore you have points (0, -1/2) and (4,-8)
two point formula: -8 + 1/2 / 4-0 = -15/8 (gradient)
y + 1/2 = -15/8 (x - 0)
y = -15/8 x - 1/2

I think this is completely wrong xDDD
I seriously cant remember how to do it, sorry D:
gta check with someone else

 

[MATHSCAPE]

[LEFT
13a)
Point M(4,-8)lies on the parabola x^2 = -2y .Find the equation of the focal chord through M.
[/LEFT]

x^2 = -2y
a = -1/2
focus (0,-1/2)
using two point formula with (0,-1/2) and (4,-8)
y+1/2 / x = -8+1/2 / 4
8y + 4 = -15x
y = -15x/8 - 1/2

answer above is correct.