Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2009 | January | Q#11

Question

The curve C has equation

,

The point P on C has x-coordinate equal to 2.

a.   Show that the equation of the tangent to C at the point P is y = 1 – 2x.

b.   Find an equation of the normal to C at the point P.

The tangent at P meets the x-axis at A and the normal at P meets the x-axis at B.

c.   Find the area of triangle APB.

Solution

a.
 

We are required to find the equation of the tangent to the curve at the point P(2,y).

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 coordinates of a point on the normal P(2,y).

Therefore, first we need y-coordinate of the point P.

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve  (or the line).

We are given that point P lies on the curve C which has equation;

We substitute x-coordinate of point P in the equation of the curve C;

Hence, coordinates of point P are (2,-3).

Next, we need slope of the tangent 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(2,-3) 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 P(2,-3).

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; 

Therefore, we need to find .

We are given equation of the curve as;

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:

We need to find the gradient of the curve C at points P(2,-3).

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 above expression for gradient of the given curve as;

For gradient of the curve at points P(2,-3), substitute  in derivative of the equation of the curve. 

Hence, gradient of the curve C at points P(2,-3) is;

Therefore, tangents to the curve at this point will also have same slope.

With coordinates of a point on the tangent P(2,-3) 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 required to find the equation of the normal to the curve at the point P(2,-3).

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 coordinates of a point on the normal P(2,-3). Therefore, we need slope of the  normal to write its equation.

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 normal to the curve at point P(2,-3) if we have slope of curve at the same point.

We have already found in (b) slope of the curve at point P(2,-3)

Now we can find slope of normal to the curve at point P(2,-3).

Now we can write equation of the normal.

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

 

c.
 

We are required to find the area of triangle APB.

Expression for the area of the triangle is;

We need both base and height of triangle APB to find area.

We need to locate the points A, B and P to sketch the triangle APB.

For this we need coordinates of points A and B while we already have, from (a), coordinates of point  P(2,-3).

We are given that points A and B are x-intercepts of tangent and normal to the curve point P. 

Since both points A and B are x-intercepts;

We need to find x-coordinates of both points A and B.

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

We have equations of both tangent and normal to the curve at point P from (a) and (b).

y-coordinate of x-intercept of tangent;

y-coordinate of x-intercept of normal;

Therefore, coordinates of points A and B are  and , respectively.

We locate points A, B and P and sketch the triangle APB as shown below.

It is evident from the diagram that;

Base of Triangle ABP=AB

Height of Triangle ABP=PC

We need to find these heights to calculate area of triangle ABP.

Expression to find distance between two given points  and is:

Therefore, we need coordinates of all points A, B, P and Q.

We have already found coordinates of points A and B as  and , respectively. 

We have also found coordinates of point P in (a).

From the diagram it is evident that vertical distance PC is 3.

Therefore;

Height of Triangle ABP=PC=3

For Base of Triangle ABP=AB;

Now we can find area of triangle ABP.

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