Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | Oct-Nov | (P1-9709/13) | Q#11
The diagram shows the curve and points A(1,0) and B(5,2) lying on the curve.
i. Find the equation of the line AB, giving your answer in the form y=mx+c.
ii. Find, showing all necessary working, the equation of the tangent to the curve which is parallel
iii. Find the perpendicular distance between the line AB and the tangent parallel to AB. Give your answer correct to 2 decimal places.
We are given points A(1,0) and B(5,2) and we are required to find equation of line AB.
Two-Point form of the equation of the line is;
Therefore, equation of line AB can be written as;
We are required to find the equation of the tangent to the curve which is parallel to AB.
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 have neither coordinates of a point on the tangent nor slope of the tangent.
But we are given that tangent is parallel to line AB.
If two lines are parallel to each other, then their slopes and are equal;
Therefore, the slope of line AB and tangent is the same.
We have found equation of line AB in (i) as;
Slope-Intercept form of the equation of the line;
Where is the slope of the line.
Therefore, slope of line AB is;
Now we need coordinates of the point on the tangent to the curve. This is the point where tangent meets the curve.
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, slope of the curve at the point where tangent meets the curve, is equal to slope of the tangent. Hence;
Therefore, we need to find the coordinates of the point on curve where slope (gradient) of the curve is .
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
We are given equation of the curve as;
Rule for differentiation of is:
We can equate this gradient of the curve with the gradient of the curve where tangent meets the curve.
To find y-coordinate of this point on the curve, we substitute value of x-coordinate of in the equation of the curve.
Hence, tangent meets the curve at point and this point also lies on the tangent.
Now we can write equation of tangent to the curve.
Point-Slope form of the equation of the line is;
We are required to find the perpendicular distance between the line AB and the tangent parallel to AB.
We have found in (i) equation of line AB as;
We have found in (ii) equation of tangent to the curve as;
It is evident that y-intercepts of line AB and tangent are;
It is also evident that slopes of both line AB and tangent are;
Expression for perpendicular distance between two parallel lines with equations and (slope of both lines being same because these are parallel) is;
Since distance cannot be negative;