Porisma

Worked solution

Find a tangent, normal and exact triangle area

A curve has x=3cos⁡θx=3\cos\theta, y=2+sin⁡θy=2+\sin\theta, 0≤θ≤2π0\le\theta\le2\pi.

(a) Show that the tangent at PP, where θ=π/6\theta=\pi/6, is y+(3/3)x−4=0y+(\sqrt3/3)x-4=0.

(b) Find the normal at PP.

(c) The tangent and normal meet the y-axis at TT and NN respectively. Find the exact area of triangle PNTPNT.

[2008/SAJC/Promo/12]

CJC · Discussion · Question 4

Answer

(a) y+(3/3)x−4=0y+(\sqrt3/3)x-4=0. (b) y=3x−2y=\sqrt3x-2. (c) Area 93/29\sqrt3/2 square units.

(a) Differentiate with respect to the parameter

dx/dθ=−3sin⁡θdx/d\theta=-3\sin\theta and dy/dθ=cos⁡θdy/d\theta=\cos\theta, so dy/dx=−cos⁡θ/(3sin⁡θ)=−1/(3tan⁡θ)dy/dx=-\cos\theta/(3\sin\theta)=-1/(3\tan\theta). At θ=π/6\theta=\pi/6, P=(33/2,5/2)P=(3\sqrt3/2,5/2) and the gradient is −3/3-\sqrt3/3. Thus y−5/2=−(3/3)(x−33/2)y-5/2=-(\sqrt3/3)(x-3\sqrt3/2), giving y+(3/3)x−4=0y+(\sqrt3/3)x-4=0.

How did this part go?

(b) Take the negative reciprocal gradient

The normal gradient is 3/3=33/\sqrt3=\sqrt3. Hence y−5/2=3(x−33/2)y-5/2=\sqrt3(x-3\sqrt3/2), or y=3x−2y=\sqrt3x-2.

How did this part go?

(c) Use the vertical base TN

Set x=0x=0 in the tangent and normal: T=(0,4)T=(0,4) and N=(0,−2)N=(0,-2). Thus TN=6TN=6, and the perpendicular height from PP to the y-axis is 33/23\sqrt3/2. The area is 12(6)(33/2)=93/2\tfrac12(6)(3\sqrt3/2)=9\sqrt3/2 square units.

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Triangle PNT has vertical base TN on the y-axis from N(0, minus 2) to T(0,4), and P(3 square root 3/2,5/2) to the right. TP is the tangent and NP the normal; the horizontal height from P to TN is 3 square root 3/2.
Triangle PNT has vertical base TN on the y-axis from N(0, minus 2) to T(0,4), and P(3 square root 3/2,5/2) to the right. TP is the tangent and NP the normal; the horizontal height from P to TN is 3 square root 3/2.
Triangle PNT has vertical base TN on the y-axis from N(0, minus 2) to T(0,4), and P(3 square root 3/2,5/2) to the right. TP is the tangent and NP the normal; the horizontal height from P to TN is 3 square root 3/2.

Common mistakes

The triangle height is the horizontal distance from P to the y-axis.

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