# Question #05cf9

##### 1 Answer

The given equation is incorrect - it is not actually the equation of a straight line ! The

# y + tx= 4t^2 + 8t#

#### 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

The equation of the given parabola is:

#y^2=16x#

Firstly let us verify that

# x=4t^2 => y^2=16x=16(4t^2) = 64t^2 = (8t)^2 \ \ \ \# QED

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

# \ 2ydy/dx=16 #

# :. dy/dx = 8/y #

So at the point

# dy/dx = 8/(8t) = 1/t #

Thus, the gradient of the tangent at

So the normal passes through

so using the point/slope form of a straight line

# y-8t = -t(x-4t^2) #

# :. y-8t = -tx+4t^2 #

# :. y + tx= 4t^2 + 8t\ \ \ # QED