# Question #05cf9

Feb 14, 2017

The given equation is incorrect - it is not actually the equation of a straight line ! The $x y$ should be $t x$ which then gives the equation as:

$y + t x = 4 {t}^{2} + 8 t$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is $- 1$

The equation of the given parabola is:

${y}^{2} = 16 x$

Firstly let us verify that $C \left(4 {t}^{2} , 8 t\right)$ does indeed lie on the parabola:

$x = 4 {t}^{2} \implies {y}^{2} = 16 x = 16 \left(4 {t}^{2}\right) = 64 {t}^{2} = {\left(8 t\right)}^{2} \setminus \setminus \setminus \setminus$ QED

Now we differentiate (implicitly) the equation of the parabola:

$\setminus 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 16$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{y}$

So at the point $C \left(4 {t}^{2} , 8 t\right)$, the value of the derivative is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{8 t} = \frac{1}{t}$

Thus, the gradient of the tangent at $C \left(4 {t}^{2} , 8 t\right)$ is ${m}_{T} = \frac{1}{t}$, and therefore the gradient of the normal is ${m}_{N} = - \frac{1}{m} _ T = - t$

So the normal passes through $C \left(4 {t}^{2} , 8 t\right)$ and has gradient $- t$,
so using the point/slope form of a straight line $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek is;

$y - 8 t = - t \left(x - 4 {t}^{2}\right)$
$\therefore y - 8 t = - t x + 4 {t}^{2}$
$\therefore y + t x = 4 {t}^{2} + 8 t \setminus \setminus \setminus$ QED