Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/12) | Q#7

Hits: 192


A curve is such that   . The line  is the normal to the curve at the point  on the curve. Given that the x-coordinate of  is positive, find

    i.       the coordinates of P,

   ii.       the equation of the curve.



We are given that equation of the line is;

We can rearrange the equation of the line as;

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, slope of the given line, by comparing both equations, is;

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;

Since, we are given that line is normal to the curve at point , slope of the curve at point  is; 

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Hence, for the given curve;

We are given that the x-coordinate of  is positive, therefore;

Corresponding values of y coordinate can be found by substituting values of x in any of the two equation i.e either equation of the line or equation of the curve.

We choose equation of the line;

Substitution of ;

Hence coordinates of point .


We can find equation of the curve from its derivative through integration;

We are given that;


Rule for integration of  is:

Rule for integration of  is:

Rule for integration of  is:

If a point   lies on the curve , we can find out value of . We substitute values of  and   in the equation obtained from integration of the derivative of the curve i.e. .

From (i), we have the coordinates of point  on the curve. Hence;

Therefore, equation of the curve is;

Please follow and like us: