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Tangents & Normals

Recall that the equation of a straight line is given by y=mx+c where m is the gradient and c is the y-intercept. Since tangents are straight lines, they will also take this form. The TANGENT to a curve at a given point has the same gradient as the curve at that point and the y-intercept can be found using the coordinates of the point.

ExampleFind the equation of the tangent to y=x^2 at the point (3,9).
A tangent is a straight line and its equation will take the form y=mx+c where we need to specify the gradient, m, and the y-intercept, c. The gradient of the tangent is the same as that of the curve and so we must first find the gradient of the curve by differentiating. The derivative is given by \frac{dy}{dx}=2x and so the gradient when x=3 is 6, i.e. m=6 and so far we have that the equation of the tangent is y=6x+c. To find c, we use the fact that the point (3,9) is on the tangent as well as the curve (it is the only point) and so when x is 3, we should get y=9: 9=6\times 3+c. This gives c=-9 and the equation of the tangent is therefore y=6x-9.

The NORMAL to a point on a curve is the line that intersects the curve at that point and is perpendicular to the curve at that point. It follows that the normal is perpendicular to the tangent at the given point. If the tangent has a gradient of m, the normal has a gradient of -\frac{1}{m}. The y-intercept of the normal, which is different from that of the tangent, can also be found using the coordinates of the given point.

ExampleFind the equation of the normal to y=x^2 at the point (3,9).
A normal is also a straight line, perpendicular to the tangent, and its equation will take the form y=mx+c. We saw in the above example that the gradient of the tangent is 6, and so the gradient of the normal is -1/6. i.e. m=-1/6. So far we have that the equation of the normal is y=-\frac{1}{6}x+c. To find c, we use the fact that the point (3,9) is on the normal as well as the curve (the only point that the curve, tangent and normal have in common) and so when x is 3, we should get y=9: 9=-\frac{1}{6}\times 3+c. This gives c=9.5 or 19/2 and the equation of the tangent is therefore y=-\frac{1}{6}x+\frac{19}{2}.