# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663A/01) | Year 2014 | January | Q#10

Question

The curve C has equation .

The point P, which lies on C, has coordinates (2, 1).

a.   Show that an equation of the tangent to C at the point P is y = 3x – 5.

The point Q also lies on C.

Given that the tangent to C at Q is parallel to the tangent to C at P,

b.   find the coordinates of the point Q.

Solution

a.

We are required to find the equation of tangent to the curve C at point .

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).

Next, 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 , then we can find slope of the tangent  to the curve at this point.

We need to find the gradient of the curve C at point .

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; Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is: 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 , substitute in derivative of the equation of the curve.     Therefore; With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.

Point-Slope form of the equation of the line is;      b.

We are given that the tangent to C at the point Q is parallel to the tangent to C at P.

If two lines are parallel to each other, then their slopes and are equal; Therefore; We first find slope of the tangent to C at P.

We have found equation of the tangent to C at P in (a) as; Slope-Intercept form of the equation of the line; Where is the slope of the line.

Therefore; Since tangent to C at P is parallel to the tangent to the curve at point Q, the gradient of the curve at  point Q must be the same. 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; 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 have found in (a) that expression for slope of the given curve is; Therefore;  We can solve this equation to find the x-coordinate of point Q.      Now we have two options.       Two values of x indicate that there are two points with same gradient on the curve.

We already know that P(2,1), therefore, point with must be the point Q.

With x-coordinate of a point on the line (or the curve) at hand, we can find the y-coordinate of the  point by substituting value of x-coordinate of the point equation of the line (or the curve).

We have equation of the curve; Substitute ;    Hence, coordinates of the point Q are ( , ).