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Exercise 6.5 · Q25

Q.The equation of one of the lines represented by the equation x2+2xycot⁡θ−y2=0x^2+2xy\cot\theta-y^2=0 is

(1) x−ycot⁡θ=0x-y\cot\theta=0
(2) x+ytan⁡θ=0x+y\tan\theta=0
(3) xcos⁡θ+y(sin⁡θ+1)=0x\cos\theta+y(\sin\theta+1)=0
(4) xsin⁡θ+y(cos⁡θ+1)=0x\sin\theta+y(\cos\theta+1)=0
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Solve m2+2mcot⁡θ−1=0m^2+2m\cot\theta-1=0 for m=x/ym=x/y to get m=−cot⁡θ±csc⁡θm=-\cot\theta\pm\csc\theta; the root −cot⁡θ−csc⁡θ-\cot\theta-\csc\theta gives the line xsin⁡θ+y(cos⁡θ+1)=0x\sin\theta+y(\cos\theta+1)=0.

Step 1. Reduce to a quadratic in m=x/ym=x/y. Divide x2+2xycot⁡θ−y2=0x^2+2xy\cot\theta-y^2=0 throughout by y2y^2:

(xy)2+2cot⁡θ(xy)−1=0⟹m2+2mcot⁡θ−1=0,m=xy\left(\dfrac{x}{y}\right)^2+2\cot\theta\left(\dfrac{x}{y}\right)-1=0 \quad\Longrightarrow\quad m^2+2m\cot\theta-1=0, \qquad m=\dfrac{x}{y}

Step 2. Solve the quadratic for mm.

m=−2cot⁡θ±4cot⁡2θ+42=−cot⁡θ±cot⁡2θ+1=−cot⁡θ±csc⁡θm=\dfrac{-2\cot\theta\pm\sqrt{4\cot^2\theta+4}}{2}=-\cot\theta\pm\sqrt{\cot^2\theta+1}=-\cot\theta\pm\csc\theta

using the identity cot⁡2θ+1=csc⁡2θ\cot^2\theta+1=\csc^2\theta. So the two roots are

m1=−cot⁡θ+csc⁡θ=1−cos⁡θsin⁡θ,m2=−cot⁡θ−csc⁡θ=−1+cos⁡θsin⁡θm_1=-\cot\theta+\csc\theta=\dfrac{1-\cos\theta}{\sin\theta}, \qquad m_2=-\cot\theta-\csc\theta=-\dfrac{1+\cos\theta}{\sin\theta}

Step 3. Convert the second root into a line equation. Each root m=x/ym=x/y corresponds to the line x−my=0x-my=0. For m2=−1+cos⁡θsin⁡θm_2=-\dfrac{1+\cos\theta}{\sin\theta}:

x−(−1+cos⁡θsin⁡θ)y=0  ⟹  x+(1+cos⁡θ)sin⁡θy=0x-\left(-\dfrac{1+\cos\theta}{\sin\theta}\right)y=0 \implies x+\dfrac{(1+\cos\theta)}{\sin\theta}y=0

Multiplying through by sin⁡θ\sin\theta (positive for 0<θ<π0<\theta<\pi):

xsin⁡θ+y(1+cos⁡θ)=0  ⟹  xsin⁡θ+y(cos⁡θ+1)=0x\sin\theta+y(1+\cos\theta)=0 \implies x\sin\theta+y(\cos\theta+1)=0

which is exactly option (4). …

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