Question

Equation, solvable for p
solve p²+ 2py cotx - y² = 0​

Answer 1

Given:

Equation, solvable for p   p² + 2py cotx - y² = 0

To find:

Solve p² + 2py cotx - y² = 0

Solution:

From given, we have a quadratic equation,

p² + 2py cotx - y² = 0

the general form of a quadratic equation is given by, ax² + bx + c = 0

The roots of this equation is given by, $x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Comparing the given quadratic equation with the standard equation, we have,

a = 1, b = 2y cotx, c = -y² and p = x.

Now, we need to find the roots of the equation using the above formula,

$p=\dfrac{-2y\cot \left(x\right)+\sqrt{\left(2y\cot \left(x\right)\right)^2-4\cdot \:1\left(-y^2\right)}}{2\cdot \:1}\\\\=\dfrac{-2y\cot \left(x\right)+\sqrt{\left(2y\cot \left(x\right)\right)^2+4\cdot \:1\cdot \:y^2}}{2\cdot \:1}\\\\=\dfrac{-2y\cot \left(x\right)+\sqrt{4y^2+\left(2y\cot \left(x\right)\right)^2}}{2\cdot \:1}\\\\=\dfrac{-2y\cot \left(x\right)+\sqrt{4y^2+\left(2y\cot \left(x\right)\right)^2}}{2}\\\\=\dfrac{2y\left(-\cot \left(x\right)+\sqrt{\cot ^2\left(x\right)+1}\right)}{2}$

$p=y\left(\sqrt{\cot ^2\left(x\right)+1}-\cot \left(x\right)\right)$

$p=\dfrac{-2y\cot \left(x\right)-\sqrt{\left(2y\cot \left(x\right)\right)^2-4\cdot \:1\left(-y^2\right)}}{2\cdot \:1}$$=\dfrac{-2y\cot \left(x\right)-\sqrt{\left(2y\cot \left(x\right)\right)^2+4\cdot \:1\cdot \:y^2}}{2\cdot \:1}$

$=\dfrac{-2y\cot \left(x\right)-\sqrt{4y^2+\left(2y\cot \left(x\right)\right)^2}}{2\cdot \:1}$

$=\dfrac{-2y\cot \left(x\right)-2y\sqrt{\cot ^2\left(x\right)+1}}{2}$

$=-\dfrac{2y\left(\cot \left(x\right)+\sqrt{\cot ^2\left(x\right)+1}\right)}{2}$

$p=-y\left(\cot \left(x\right)+\sqrt{\cot ^2\left(x\right)+1}\right)$

Therefore, the values of p are:

$p=y\left(\sqrt{\cot ^2\left(x\right)+1}-\cot \left(x\right)\right),\:p=-y\left(\cot \left(x\right)+\sqrt{\cot ^2\left(x\right)+1}\right)$

Answer 2

(y-c1sec square (x/2))+(y-c2cosec square (x/2)