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Except you're not getting any p00n.

$Let $ Q $ have co-ordinates $ (x,y).\\ $Since $ F $ divides $ PQ $ into the ratio $ t:\frac{1}{t}, $ we get that, in $ x,\\ 0 = \dfrac{tx+\frac{1}{t}\cdot 2t}{t+\frac{1}{t}} \Rightarrow tx +2 =0 \Rightarrow x = -\dfrac{2}{t}. \\ $In $y, 1 = \dfrac{ty+\frac{1}{t}\cdot t^2}{t+\frac{1}{t}}...

OzKo was right - errday 4U exam errdayy

Selective is overrated kthxbai.

Yeah some juggling results in using e := \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n

Well if it is a necessity you have no choice...

You can do that but perhaps need some inverse function motivation...

I mentioned that above.

Sometimes when you end up with an expression in the form D^2(x) = f(x) one should remember that one need not implicitly differentiate or take a square root when attempting to find stationary points of D(x). This is because D(x) = \sqrt{f(x)} \Rightarrow \dfrac{dD}{dx} =...

You can use any valid HSC method in any paper - the thing is - you will never need to. E.g. integration by parts of y = lnx when you can use the inverse function (x = lny and integrate wrt y)

Alternatively y^2 = 4ax \Rightarrow y = \sqrt{4a} \sqrt{x} (y $ positive) so $ \dfrac{dy}{dx} = \sqrt{4a}\cdot \dfrac{1}{2\sqrt{x}}\\ $But remember that $ \sqrt{x} = \dfrac{y}{\sqrt{4a}} \Rightarrow \dfrac{dy}{dx} = \sqrt{4a}\cdot \dfrac{1}{2\frac{y}{\sqrt{4a}}} \\ = \dfrac{2a}{y}.

Why does this thread exist? Wow.

You two can't keep it in your pants for 2 seconds.

xx_0 = 2a(y+y_0) $ and $ x^2=4ay \Rightarrow y = \dfrac{x^2}{4a}. $ The chord of contact meets the parabola in the equation $\\ xx_0 = 2a\left(\dfrac{x^2}{4a}+y_0\right). This is just finding the x coordinates of the intersection between a line and a parabola e.g. $Find a quadratic in $ x $...

How do I become you?

Asians.

This is a 4U Question not a 3U one.

You dont have to but you must never work both sides simultaneously and end with 0=0 or something. You can do tan(lhs)=.... And then separately, tan(rhs)=.......=tan(lhs) and since lhs and rhs are acute, lhs=rhs

Still having trouble?