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
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Question
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.
Solution
i.
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 .
ii.
We can find equation of the curve from its derivative through integration;
We are given that;
Therefore;
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;
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