Hey, I have no clue where "57p - 102q = 624" came from (in the mark scheme). I understand everything else, except that. Can someone please explain this for me?
#A Level: Implicit Differentiation
gray hare
outer plover
implied by this
conclude that
26(q+24) = 19(3p-4q)
rearrange to the thing you're given
@gray hare
soft mesa
Hey, I have no clue where "57p - 102q = 624" came from (in the mark scheme). I u...
Alright, let's see.
px^3 + qxy + 3y^2 = 26
We take the differential.
3px^2 dx + q(xdy + ydx) + 6y dy = 0
Now, we express dy/dx.
(3px^2 + qy)dx + (qx + 6y)dy = 0
dy/dx = (-3px^2 - qy)/(qx + 6y)
So that's correct.
Next, we know that P = (-1, -4) belongs to our curve. So, let's substitute those coordinates.
-p + 4q + 48 = 26
p - 4q = 22
That's one equation for p and q.
Next, we know that the normal at P is 19x + 26y + 123 = 0. This line has slope -19/26.
The gradient of the normal is -dx/dy. So:
-dx/dy = (qx + 6y)/(3px^2 + qy) = -19/26
We also know that this is a normal at P = (-1, -4), so let's substitute those coordinates.
(-q - 24)/(3p - 4q) = -19/26
19(3p - 4q) = 26(q + 24)
19p - 34q = 208
So, we get a system of two equations:
p - 4q = 22
19p - 34q = 208
The equation that they obtained is just three times the second equation.
gray hare
26(q+24) = 19(3p-4q)
The sad part is I literally did this... But I added a q
As in +6(-4)... = -24q
So, after re-arranging I ended with Ap + Bq = 0 (where A and B are constants)
But now I understand, I'm just an idiot - thanks guys