Areas from Tangents and Normals | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Find all three vertices first

When a triangle has no horizontal or vertical side, a coordinate difference is not its perpendicular height. Draw a surrounding rectangle instead: subtract the outside right-angled triangles.

Take $y=(4-\sqrt{x})^3$ . Its normals at $P=(4,8)$ and $Q=(9,1)$ meet at $R$ . Use the chain rule from the earlier lesson:

y'=-\frac{3(4-\sqrt{x})^2}{2\sqrt{x}},\qquad x>0.

At $x=4,9$ , the tangent gradients are $-3,-1/2$ , so the normal gradients are $1/3,2$ . Use each gradient with its own point:

\begin{aligned}y-8&=\tfrac13(x-4),\\3y&=x+20;\\y-1&=2(x-9),\\y&=2x-17.\end{aligned}

Substitute the second equation into the first:

\begin{aligned}3(2x-17)&=x+20,\\5x&=71,\\x&=14.2,\\y&=11.4.\end{aligned}

Thus $R=(14.2,11.4)$ ; it satisfies both normal equations.

Subtract the outside triangles

P = (4,8), Q = (9,1), R = (14.2,11.4). Rectangle edges are horizontal or vertical.

Horizontal lengths are differences of x-coordinates; vertical lengths are differences of y-coordinates. The rectangle has width $14.2-4=10.2$ and height $11.4-1=10.4$ . Each outside triangle uses two perpendicular rectangle edges:

$A_1,A_2,A_3$ are the areas numbered in the diagram.

\begin{aligned}A_1&=\tfrac12(5)(7)=17.5,\\A_2&=\tfrac12(5.2)(10.4)=27.04,\\A_3&=\tfrac12(10.2)(3.4)=17.34.\end{aligned}

The outside areas total $61.88$ . Subtract from the rectangle:

\begin{aligned}\text{Area}&=106.08-61.88\\&=44.2\text{ units}^2.\end{aligned}

The result is positive and smaller than the rectangle's area.

Use different normals

Your turnIndependent practice

The normal at $P=(1,3)$ is $4y=x+11$ . The normal at $Q=(3,1)$ is $2y=3x-7$ . They meet at $R$ . Find $R$ and the area of triangle $PQR$ . Draw the triangle before calculating.

Show worked answer

Equate the expressions for $4y$ : $x+11=6x-14$ , so $R=(5,4)$ . A surrounding rectangle has width $5-1=4$ and height $4-1=3$ . Subtract the lower-left, lower-right and upper triangles:

\begin{aligned}A_1&=\tfrac12(2)(2)=2,\\A_2&=\tfrac12(2)(3)=3,\\A_3&=\tfrac12(4)(1)=2.\end{aligned}

\begin{aligned}\text{Area}&=12-2-3-2\\&=5\text{ units}^2.\end{aligned}