Question

Locus of point of intersection of the lines x sinetheta - y costheta = 0 and ax sectheta- b ycosectheta = a²-b²

plz answer me whole method​

Answer 1

Step-by-step explanation:

Given lines are

$\tt \: x \: sin( \theta) - y \: cos (\theta) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt ax \: sec (\theta) - by \: cosec (\theta) = {a}^{2} - {b}^{2}$

Now,

From 1st equation,

$\tt \: x = y \: cot( \theta)$

put this in 2nd equation,

$\tt ay \: cot (\theta) \: sec (\theta) - by \: cosec (\theta) = {a}^{2} - {b}^{2}$

$\tt \implies \: ay \: cosec (\theta) - by \: cosec (\theta) = {a}^{2} - {b}^{2}$

$\tt \implies \: (a - b)y \: cosec (\theta) = {a}^{2} - {b}^{2}$

$\tt \implies \: y \: cosec (\theta) = a + b$

$\tt \implies \: y = (a + b)sin (\theta)$

So,

$\tt \: x = (a + b) sin( \theta) cot( \theta)$

$\tt \: \implies \: x = (a + b) cos( \theta)$

Required locus

$\sf\large \color{red} {x}^{2} + {y}^{2} = (a + b)^{2}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:xsin\theta - ycos\theta = 0$

$\rm :\longmapsto\:xsin\theta = ycos\theta$

$\rm :\longmapsto\:x = \dfrac{ycos\theta }{sin\theta } - - - - (1)$

Also, given that,

$\rm :\longmapsto\:axsec\theta - bycosec\theta = {a}^{2} - {b}^{2}$

On substituting the value of x, we get

$\rm :\longmapsto\:a \times \dfrac{ycos\theta }{sin\theta } \times sec\theta - bycosec\theta = {a}^{2} - {b}^{2}$

We know,

$\boxed{ \bf{ \:sec\theta = \frac{1}{cos\theta }}}$

So, on substituting this, value we get

$\rm :\longmapsto\:a \times \dfrac{ycos\theta }{sin\theta } \times \dfrac{1}{cos\theta } - bycosec\theta = {a}^{2} - {b}^{2}$

$\rm :\longmapsto\:a \times \dfrac{y}{sin\theta } - bycosec\theta = {a}^{2} - {b}^{2}$

$\rm :\longmapsto\:a \times \dfrac{y}{sin\theta } - by \times \dfrac{1}{sin\theta } = {a}^{2} - {b}^{2}$

$\rm :\longmapsto\:\dfrac{y}{sin\theta }(a - b) = {a}^{2} - {b}^{2}$

$\rm :\longmapsto\:\dfrac{y}{sin\theta }(a - b) = (a - b)(a + b)$

$\rm :\longmapsto\:\dfrac{y}{sin\theta }= a + b$

$\bf\implies \:sin\theta = \dfrac{y}{a + b} - - - - (2)$

On substituting the value in equation (1),

$\rm :\longmapsto\:x = \dfrac{ycos\theta (a + b)}{y }$

$\rm :\longmapsto\:x = (a + b)cos\theta$

$\bf\implies \:cos\theta = \dfrac{x}{a + b} - - - - (3)$

On squaring equation (2) and (3),and add we get

$\rm :\longmapsto\: {sin}^{2}\theta + {cos}^{2}\theta = {\bigg[\dfrac{x}{a + b} \bigg]}^{2} + {\bigg[\dfrac{y}{a + b} \bigg]}^{2}$

$\bf\implies \: {\bigg[\dfrac{x}{a + b} \bigg]}^{2} + {\bigg[\dfrac{y}{a + b} \bigg]}^{2} = 1$

OR

$\bf\implies \: {x}^{2} + {y}^{2} = {(a + b)}^{2}$

• This is the required locus.