Question: Find the separate equations of straight lines represented by the following equation: x² + 2xy cosec\alpha + y² = 0.


Solution:
Given

Single equation of pair of straight lines is
x² + 2 xy cosec\alpha + y² = 0

[ Dividing by y² ]

or, \dfrac{x²}{y²} + \dfrac{2\; xy \;cosec\alpha}{y²} + \dfrac{y²}{y²} = 0

or, \left ( \dfrac{x}{y} \right )^2  + 2 \;cosec\alpha \dfrac{x}{y} + 1 = 0 - (i)

Comparing equation (i) with ax² + bx + c = 0
We get,
a = 1, b = 2 cosec\alpha, c = 1, x = x/y


Using formula of quadratic equation,

or, \dfrac{x}{y} = \dfrac{-b \pm \sqrt{b² - 4ac}}{2a}

= \dfrac{- 2 cosec\;\alpha \pm \sqrt{(2cosec\alpha)² - 4×1×1}}{2}

= \dfrac{- 2 cosec\;\alpha \pm \sqrt{4cosec²\alpha - 4}}{2}

= \dfrac{- 2 cosec\;\alpha \pm \sqrt{4(cosec²\alpha - 1)}}{2}

= \dfrac{- 2 cosec\;\alpha \pm 2\sqrt{cosec²\alpha - 1}}{2}

[ cosec² A - 1 = cot² A ]

= \dfrac{-2 cosec\alpha \pm 2\sqrt{cot² \alpha}}{2}

or, \dfrac{x}{y} = \dfrac{- 2 cosec\alpha \pm 2cot \alpha}{2}


Taking positive sign,

or, \dfrac{x}{y} = \dfrac{- 2 cosec\alpha + 2cot \alpha}{2}

or, \dfrac{x}{y} = \dfrac{-2 (cosec\alpha - cot\alpha)}{2}

or, x = -y(cosec \alpha -cot\alpha)

or, x = - ycosec \alpha + x cot\alpha

or, x + ycosec \alpha - ycot\alpha = 0

Taking negative sign,

or, \dfrac{x}{y} = \dfrac{- 2 cosec\alpha - 2cot \alpha}{2}

or, \dfrac{x}{y} = \dfrac{-2 (cosec\alpha + cot\alpha)}{2}

or, x = -y(cosec\alpha +cot\alpha)

or, x = -ycosec\alpha -ycot\alpha

or, x + ycosec\alpha + ycot\alpha = 0


Hence, the required separate equations of straight lines represented by the above equation is (x + ycosec \alpha - ycot\alpha = 0) and (x + ycosec\alpha + ycot\alpha = 0).


Related Notes and Solutions:

Here is the website link to all the important formulae of Coordinate Geometry of Class 10.

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