# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2015 | June | Q#10

Hits: 148

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;         