Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2013 | June | Q#11

Question

Figure shows a sketch of the curve H with equation;

a. Give the coordinates of the point where H crosses the x-axis.

b. Give the equations of the asymptotes to H.

c. Find an equation for the normal to H at the point P(–3, 3).

This normal crosses the x-axis at A and the y-axis at B.

d. Find the length of the line segment AB. Give your answer as a surd.

Solution

We are given equation of curve H;

We are required to find the coordinates of the point where the curve meets the x-axis.

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 substitute y=0 in equation of the curve;

Therefore, coordinates of x-intercept of the curve H are .

We are given that;

Next we are required to state the equations of asymptotes.

An asymptote is a line that a curve approaches, as it heads towards infinity. Both horizontal and vertical asymptotes may exist for a given graph. The distance between the curve and the asymptote tends to zero as they head to infinity.

For the graph sketched above, we can draw horizontal and vertical asymptotes as shown below.

desmos-graph (4).png

It is evident that horizontal asymptote (orange line) can be stated with equation;

Similarly, vertical asymptote (blue line) can be stated with equation;

We are required to find the equation of the normal to the curve at the point P(-3,3).

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

Therefore, we need to find gradient of the curve H at point P(-3,3).

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:

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

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(-3,3), substitute in derivative of the equation of the curve.

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

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

Now we can write equation of the normal.

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

We are given that normal to the curve H at point P(-3,3), whose equation as follows, has A and B points as x and y intercepts, respectively.

We are required to find length of AB.

Expression for the distance between two given points and is:

Therefore, we need coordinates of both A and B points.

First we find coordinates of point A which is x-intercept of the normal to the curve H.

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 substitute y=0 in equation of the normal;

Therefore, coordinates of A are (-4,0).

Next we find coordinates of point B which is y-intercept of the normal to the curve H.

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