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Exercise 6.1 · Q9

Q.The coordinates of a moving point PP are (a2(csc⁡θ+sin⁡θ), b2(csc⁡θ−sin⁡θ))\left(\dfrac{a}{2}\left(\csc\theta + \sin\theta\right),\ \dfrac{b}{2}\left(\csc\theta - \sin\theta\right)\right), where θ\theta is a variable parameter. Show that the equation of the locus of PP is b2x2−a2y2=a2b2b^2x^2 - a^2y^2 = a^2b^2.

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Add and subtract the two coordinate equations to isolate csc⁡θ\csc\theta and sin⁡θ\sin\theta separately, then use the identity csc⁡θsin⁡θ=1\csc\theta\sin\theta=1.

Step 1. Write the coordinate equations. Let P=(x,y)P=(x,y) with

x=a2(csc⁡θ+sin⁡θ),y=b2(csc⁡θ−sin⁡θ).x=\frac{a}{2}(\csc\theta+\sin\theta), \qquad y=\frac{b}{2}(\csc\theta-\sin\theta).

Multiplying through,

2xa=csc⁡θ+sin⁡θ...(1),2yb=csc⁡θ−sin⁡θ...(2).\frac{2x}{a}=\csc\theta+\sin\theta \qquad \text{...(1)}, \qquad \frac{2y}{b}=\csc\theta-\sin\theta \qquad \text{...(2)}.

Step 2. Add (1) and (2).

2xa+2yb=2csc⁡θ  ⇒  csc⁡θ=xa+yb.\frac{2x}{a}+\frac{2y}{b}=2\csc\theta \;\Rightarrow\; \csc\theta=\frac{x}{a}+\frac{y}{b}.

Step 3. Subtract (2) from (1).

2xa−2yb=2sin⁡θ  ⇒  sin⁡θ=xa−yb.\frac{2x}{a}-\frac{2y}{b}=2\sin\theta \;\Rightarrow\; \sin\theta=\frac{x}{a}-\frac{y}{b}. …

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