# Maths ‘A’ level: now for a tough one

From the same paper as the previous two questions, here is the hardest question (as measured by marks on offer – 12):

Sketch the curve given parametrically by $x = t^2$, $y = t^3$
Show that an equation of the normal to the curve at the point $A(4,8)$ is $x + 3y - 28 = 0$
This normal meets the x-axis at the point $N$. Find the area of the region enclosed by the arc $OA$ of the curve, the line segment $AN$ and the x-axis.

While I can see the ideas behind each part of this question I am not at all sure where to begin with it.

For background, by the marking scheme you should take about 16 minutes to answer this – (and about 5 minutes per question for the earlier two). No pressure then.

## 8 thoughts on “Maths ‘A’ level: now for a tough one”

1. tobi says:

Once you have the sketch everything else becomes much easier. You get the sketch by asking yourself “Which points (x, y) can be generated by varying t?”. You notice that x is always positive and that for each x there are two y values, one with positive and one with negative t. So draw two graphs: One for t positive and one for t negative. For each of those domains transform the two equations given into the form y = f(x). That makes drawing the two curves trivial and also allows to get the derivative later.

1. Yes but it talks of one curve, not two. Or have I missed something?

1. tobi says:

Not sure. It is one curve though because the two curves meet at (0, 0).

2. OK, sketched the curve, it’s one curve with an inflection point at 0. But I make it $f(x)=x^{\frac{3}{2}}$ and equation of normal as $y=\frac{3}{2}\sqrt{x} + 2$, so plainly I have got something wrong here

1. OK so the equation for the normal is $y = \frac{3}{2}\sqrt{x} + 5$, and I can make that look like $3y - \frac{9}{4}x - 15 = 0$ and $-\frac{9}{4}x - 15 = x - 28$ at $A$. But is there a cleaner way?

2. ha ha. I now I realise I have completely misunderstood the question. U for me.

3. Isn’t this basic calculus? The reason they call it a curve is because it’s not a function of a single variable. The only reason you can pretend it is because it only asks you about one piece of the curve.