FP1 - Edexcel - Coordinate systems - Need help with question

Ah yeah, stumped me there lol. The square roots have very weird rules.

I'd just stick to implicit to avoid the confusion.


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I think that's where I am going wrong, how do we know from example 11 that t has a negative gradient? As if t was negative then wouldn't the equation I am supposed to be proving have a positive gradient. It is weird but it seems that it seems to work whatever the case.

OK, let's nail this mother

Firstly, think about the equation they start with in the question:

$y^2 = 4ax, a>0$

This defines a parabola with 2 "branches" - one above the x-axis and one below the x-axis. We are told that a>0, and y^2>0, so the graph is only defined for x>0. However, if y>0 corresponds to a point on the upper branch then there is a corresponding point on the lower branch with a negative y value since we could square both values of y and get the same number.

What they have done in the question is treat the uppermost branch of the parabola, which can be written: $y = \sqrt{4ax}$ (using the convention that the radical sign represents the positive square root). All the analysis they have done works fine because for all the points on this branch we have t>0, x>0, y>0 so we can take square roots in a "sensible" way.

If we tried to do a similar thing for the lower branch, we would be considering the function $y = -\sqrt{4ax}$.
In this case we have to be careful when we take square roots. Because t is negative on this branch of the parabola,
$\displaystyle -\frac{\sqrt{a}}{\sqrt{x}} = -\frac{\sqrt{a}}{(\sqrt{a}(-t))} = 1/t$ as before.

Note that the gradient has to be 1/t in both cases: above the x-axis the curve has a positive gradient and t>0; below the axis the curve has a negative gradient and t<0.

Now for the safer methods

Method 1 - Implicit differentiation
From $y^2 = 4ax$
we have $2y\dfrac{dy}{dx} = 4a, \dfrac{dy}{dx} = 4a/2y = 2a/y = 2a/2at = 1/t$

Method 2 - Parametric differentiation
$x = at^2, y = 2at, \dfrac{dy}{dx} = \dfrac{dy}{dt} / \dfrac{dx}{dt} = 2a/2at = 1/t$

With these approaches we can ignore square roots altogether and arrive at the desired equation for the normal

Thanks a lot you really cleared things up for me. I tried to write out the gradient with a negative t but where you put the square root of t^2 equal to -t I put it equal to t. Am I right in saying that in algebra t^2 is equal to -t and positive t.

It's because we have the convention that the square root is positive:

if t>0, $\sqrt{t^2} = t$
If t<0, $\sqrt{t^2} = -t$

Your book's approach is simplistic but works because of the symmetry of the (complete) curve so you get the same answer as if you did the analysis in detail.

Personally I would hope that you either don't get such a horrible question, or if you do that you are allowed to use implicit or parametric differentiation

I understand how to work out the normal of the equation but here is where I am confused... When you square root something there is a positive and a negative solution so why have they presumed that y is positive. Isn't there two solutions?

The square root by definition is always positive.

If you look at post 22 at what davros put at the part where he worked out the gradient for a negative branch, do you know how that works as if you leave the gradient formula in the form -1 over the square root of t^2. Then if t was -3 which gives a negative y value on the negative branch then you would expect the gradient to be negative. So substituting t= -3 then it becomes -1 over the square root of (-3)^2 which is equal to -1 over -3 which is positive 1/3 which isn't a negative gradient.

-1/t^2 if t is -3 then it's -1/9 which is negative...

I meant the square root of t^2 so it would be -3 and so 1/9 which would make the gradient positive when it needs to be negative.

$\sqrt{t^2} = |t|$ - we must always get a positive number from the square root operator.

E.g, $\sqrt{(-1)^2} = +1$

I've had another think about this question and there's another way of working out the tangent that avoids calculus altogether!

Suppose y=mx+c is a tangent to $y^2 = 4ax$ at the point $(at^2,2at)$
The line and curve intersect, so:
$(mx+c)^2 = 4ax$
or
$m^2x^2 + 2mcx + c^2 = 4ax$
so
$m^2x^2 + (2mc-4a)x + c^2 = 0$
For a tangent we need equal roots, which means the discriminant must be 0:
$(2mc-4a)^2 = 4m^2c^2$

Expanding the LHS:
$4m^2c^2 - 16mca + 16a^2 = 4m^2c^2$
or
$16a^2 = 16mca$
so $c = a/m$

Now, at $(at^2,2at)$
$y=mx+c$
so
$2at = mat^2 + c = mat^2 + (a/m)$

Hence
$2mat = m^2at^2 + a$
$a \neq 0$ so $2mt = m^2t^2 + 1$
or $m^2t^2 - 2mt + 1 = 0$
which is $(mt - 1)^2 = 0$

Therefore, $mt=1$
giving
$m=1/t, c = a/m = at$
and
$y = (1/t)x + at$
as the equation of the tangent.