Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2005 | January | Q#7

Question

The curve C has equation , . The point P on C has x-coordinate 1.

a)   Show that the value of at P is 3.

b)  Find an equation of the tangent to C at P.

This tangent meets the x-axis at the point (k,0).

c)   Find the value of k.

Solution

a)

We need to find the gradient of the curve C at point P(1,y).

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that  particular point.

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; We are given equation of the curve as; Therefore; Rule for differentiation is of is:     Rule for differentiation is of is: Rule for differentiation is of is:     For gradient of the curve at point P(1,y), substitute in derivative of the equation of the curve.    b)

We are required to find the equation of tangent to the curve C at point P(1,0).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We already have x-coordinate of a point on the tangent as P(1,y).

With x-coordinate of point of on the curve at hand, we can find the y-coordinate of the point by  substituting value of x-coordinate of the point in equation of the curve.

We are given equation of the curve as; Substituting ;   Hence coordinates of point P on the curve C are .

We need to find slope of tangent at in order to write its equation.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the  same point;  Therefore, if we can find slope of the curve C at point P(1,8) then we can find slope of the tangent to  the curve at this point.

We already have found in (a) gradient of the curve C at point P(1,8).  Therefore; With coordinates of a point on the tangent P(1,8) and its slope in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;     c)

Since tangent meets the x-axis at point (k,0), this is x-intercept of the tangent.

The point at which curve (or line) intercepts x-axis, the value of . So we can find the  value of coordinate by substituting in the equation of the curve (or line).

We have found equation of tangent in (b) as; Therefore, substituting and ;    