# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2017 | May-Jun | (P1-9709/13) | Q#6

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**Question**

The line 3y+x=25 is a normal to the curve y= x^{2} – 5x + k. Find the value of the constant k.

**Solution**

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, by finding slopes of both the line and the curve we can find the value of k.

First let’s find slope of the line.

Slope-Intercept form of the equation of the line;

Where is the slope of the line.

We are given equation of the line as;

We can rearrange this given equation in slope-intercept form as follows.

Hence, slope of the line is;

Next we find slope of the curve.

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;

Therefore for gradient of the curve;

Rule for differentiation of is:

Rule for differentiation of is:

Rule for differentiation of is:

Since k in the equation of curve is a constant;

Hence, slope of the curve is;

Therefore;

It is evident that slopes of both normal and the curve are negative reciprocal of each other only at the point where normal and curve intersect.

We have found x-coordinate of this point of intersection as .

With x-coordinate of a point of intersection of two lines (or line and the curve) at hand, we can find the y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value of x-coordinate of the point of intersection in any of the two equations;

We choose equation of line;

Substituting ;

Hence, line and curve intersect at point . Since this point also lies on the curve, we can find the value of k by substituting these coordinates in equation of the curve.

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