Question

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value of θ when these two points coincide. (0 ≤ θ < 2π)​

Answer 1

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

Given that,

• Coordinates of P is (sinθ + 2, tanθ - 2)

• Coordinates of Q is (4sin²θ + 4sinθcosθ + 2acosθ, 3sinθ - 2cosθ + a)

As it is given that, Coordinates of P and Q coincides.

So, it means

$\rm \: sin\theta +2 = 4sin^{2} \theta +4sin\theta cos\theta +2acos\theta - - (1)$

and

$\rm \: tan\theta -2 = 3sin\theta -2cos\theta +a$

can be rewritten as

$\rm \: \frac{sin\theta }{cos\theta } -2 = 3sin\theta -2cos\theta +a$

$\rm \: \frac{sin\theta - 2cos\theta }{cos\theta } = 3sin\theta -2cos\theta +a$

$\rm \: sin\theta - 2cos\theta = 3sin\theta \: cos\theta -2cos^{2} \theta +acos\theta - - (2)$

Let assume that $sin\theta = x\:and\:cos\theta=y$

So, equation (1) and (2) can be rewritten as

$\rm \: x +2 = {4x}^{2} + 4xy +2ay - - (3)$

and

$\rm \: x - 2y = 3xy -2 {y}^{2} +ay - - (4)$

On multiply equation (4) by 2, we get

$\rm \: 2x - 4y = 6xy -4{y}^{2} +2ay - - (5)$

On Subtracting equation (5) from (3), we get

$\rm \: x + 2 - 2x + 4y = {4x}^{2} + 4xy - 6xy + 4{y}^{2}$

$\rm \: 2 - x + 4y = 4({x}^{2} + {y}^{2}) - 2xy$

$\rm \: 2 - x + 4y = 4 - 2xy$

$\rm \: 2 - x + 4y - 4 + 2xy = 0$

$\rm \: - x + 4y - 2 + 2xy = 0$

$\rm \: - (x + 2) + 2y(2 +x) = 0$

$\rm \: (x + 2)( - 1 + 2y) = 0$

$\rm\implies \:x = - 2 \: \: or \: \: y \: = \: \frac{1}{2}$

$\rm\implies \:sin\theta = - 2\: \: \{rejected \} \: \: or \: \: cos\theta = \: \frac{1}{2}$

So,

$\rm\implies \:cos\theta = \dfrac{1}{2}$

$\rm\implies \:\theta = \dfrac{\pi}{3} \: \: or \: \: \dfrac{5\pi}{3}$

Now, Consider

$\boxed{ \rm{ \:\theta \: = \: \dfrac{\pi}{3} \: }}$

So,

$\rm \: tan\theta -2 = 3sin\theta -2cos\theta +a$

can be rewritten on substituting the values, we get

$\rm \: tan\dfrac{\pi}{3} -2 = 3sin\dfrac{\pi}{3} -2cos\dfrac{\pi}{3} +a$

$\rm \: \sqrt{3} -2 = 3 \times \frac{ \sqrt{3} }{2} -2 \times \frac{1}{2} +a$

$\rm \: \sqrt{3} -2 = \frac{ 3\sqrt{3} }{2} -1 +a$

$\rm \: \sqrt{3} -2 - \frac{ 3\sqrt{3} }{2} + 1 = a$

$\rm \:\bf\implies \:a \: = \: - \: \frac{ \sqrt{3} }{2} - 1$

Now, Consider

$\boxed{ \rm{ \:\theta \: = \: \dfrac{5\pi}{3} \: }}$

So,

$\rm \: tan\theta -2 = 3sin\theta -2cos\theta +a$

can be rewritten as

$\rm \: tan\dfrac{5\pi}{3} -2 = 3sin\dfrac{5\pi}{3} -2cos\dfrac{5\pi}{3} +a$

$\rm \: - \sqrt{3} -2 = - 3 \times \frac{ \sqrt{3} }{2} -2 \times \frac{1}{2} +a$

$\rm \: - \sqrt{3} -2 = - \frac{3\sqrt{3} }{2} -1 +a$

$\rm \: - \sqrt{3} -2 + \frac{ 3\sqrt{3} }{2}+1 = a$

$\rm \:\bf\implies \:a \: = \: \frac{ \sqrt{3} }{2} - 1$