|ll|lll|'s question thread

[|ll|lll|]

What point on the parabola is nearest to the point (a, 0), where 0 < p < a?

[b^3]

So to find which one is closest to it, we need to find the magnitude of the line join the two points, then the minimum of that and find the point.
So the magnitude of the distance between the two points is

This will be a miniumum when (x-a)²+2px is a minimum
so ley f(x)=(x-a)²+2px
then f'(x)=2(x-a)+2p
2(x-a)+2p=0
x=a-p
Now when x=a-p, y²=2p(a-p)
so
So or
So the point would be at and

EDIT: Something looks logically wrong here, what was the answer?

[b^3]

7 marks! I could see maybe 5 unless they take splitting the plus and minus up twice as 2 answer marks.

[brightsky]

since it's a parabola symmetrical about the x-axis, there will be two answers.
let the required points be (b, sqrt(2pb)) and (b, -sqrt(2pb))
y^2 = 2px
2yy' = 2p
y' = 2p/2y = p/y
the closest point would be when the line from point (a,0) to (b,sqrt(2pb)) and (b,-sqrt(2pb)) is perpendicular.
look at the point (b,sqrt(2pb)) WLOG:
this line has gradient -y/p = -sqrt(2pb)/p
this is equivalent to -sqrt(2pb)/(a-b)
from this we can form an equation:
sqrt(2pb)/p = sqrt(2pb)/(a-b)
p = a-b
b = a-p
so the required points are (a-p,sqrt(2p(a-p))) and (a-p,-sqrt(2p(a-p))).

[|ll|lll|]

Here's an exam 1 question

2 cos^2 (x) + 9 sin (x) = 6 for  0 <= x <= 2pi

After solving a quadratic, I have x = pi/4, 3pi/4, 5pi/4, 7pi/4

However, when I solve that in the calculator, I only have two solutions and they're not the same solutions as mine.

[tony3272]

I think you may have made a mistake in solving your quadratic. This is what i did:







by letting





or

implies that:  or

as the second one has no solution: