Review question

# Can we show that these normals meet on the $x$-axis? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5696

## Solution

A curve is given parametrically by the equations $x=4-t^2, y=4t.$

Find the equations of the normals to the curve at the points where the curve meets the $y$-axis. Show that these normals meet on the $x$-axis.

The curve meets the $y$-axis where $x=0$, hence at $t=\pm 2$. The corresponding points on the curve are $P= (0,8)$ and $Q= (0,-8)$.

Differentiating we find $\dfrac{dy}{dt} = 4$ and $\dfrac{dx}{dt} = -2t$. By the Chain Rule, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = -\dfrac{2}{t} ,$ so the gradient of the normal at the point given by the parameter $t$ is $\dfrac{t}{2}$.

The normal at $P$ is $y = x + 8$, and the normal at $Q$ is $y = -x - 8$.

Setting the two equations equal to each other gives $-x-8=x+8 \implies x=-8 \implies y=0$, and hence the lines meet on the $x$-axis.

The curve is a parabola, symmetrical about the $x$-axis. The symmetry of the situation guarantees that the normals will meet on the $x$-axis.