Question

A curve with equation y=f(x) passes through the point (4,9).

Given that

 , x > 0

a.   find f(x), giving each term in its simplest form.

Point P lies on the curve.

The normal to the curve at P is parallel to the line 2y + x = 0

b.   Find x coordinate of P.

Solution

a.
 

We are given;

We are given coordinates of a point on the curve (4,9).

We are required to find the equation of y in terms of x ie f(x).

We can find equation of the curve from its derivative through integration;

Therefore,

Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and    in the equation obtained from integration of the derivative of the curve i.e. .

Therefore, substituting the coordinates of point (4,9) in above equation;

Therefore, equation of the curve C is;

b.
 

We are given that point P lies on the curve and the normal to the curve at P is parallel to the line 2y  + x = 0.

We are required to find the x coordinate of P.

We can find the x-coordinates of P through its gradient.

If we can find the gradient o the curve at point P, we can equate it with given expression of  derivative (gradient) of the curve to find x-coordinates of point P.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Therefore, we need gradient of the curve at point P.

We are given that normal to the curve at point P has equation 2y + x = 0.

If a line  is normal to the curve , then product of their slopes  and  at that point (where line  is normal to the curve) is;

Therefore;

We can find slope of the curve at point P if we can find slope of the normal to the curve at this point. 

We are given equation of the normal to the curve at point P.

Slope-Intercept form of the equation of the line;

Where  is the slopeof the line. 

We can rearrange the given equation in slope-intercept form as follows.

Hence, slope of the normal to the curve at point P is  .

Now we can find slope of the curve at point P.

Gradient (slope)  of the curve  at a particular point  can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

Therefore;