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Question: How do you convert \({{x}^{2}}-12y-36=0\) to polar form?...

How do you convert x2−12y−36=0{{x}^{2}}-12y-36=0 to polar form?

Explanation

Solution

Conversion of cartesian to polar coordinate system requires some basic conversion formula which are
x=rcos⁡θ y=rsin⁡θ r=x2+y2 tan⁡θ=yx \begin{aligned} & x=r\cos \theta \\\ & y=r\sin \theta \\\ & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\\ & \tan \theta =\dfrac{y}{x} \\\ \end{aligned}
We replace the variables in the given equation with the polar coordinates using the first two formulas and then proceed to simplify the converted equation.

Complete step-by-step solution:
The given equation that we need to convert to polar form in this problem is,
x2−12y−36=0{{x}^{2}}-12y-36=0
After analysing the above equation, we can see that the two variables involved in the above equation are x and y. According to our knowledge, x and y variables belong to the cartesian system of coordinates. The cartesian system of coordinates incorporates two axes – the x axis and the y axis. The location of a point in the x-y plane is shown by its distance from the x and y axes respectively. The polar coordinates however, represent the location of a point on the r- θ\theta plane using the distance “r” of the point from a predefined point, called the origin and its inclination θ\theta from a predefined direction.
Now, for the conversion of cartesian to polar coordinate system has a set of formulae which are,
x=rcos⁡θ y=rsin⁡θ r=x2+y2 tan⁡θ=yx \begin{aligned} & x=r\cos \theta \\\ & y=r\sin \theta \\\ & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\\ & \tan \theta =\dfrac{y}{x} \\\ \end{aligned}
In the given equation, we replace x and y using the first two formula and get,
⇒(rcos⁡θ)2−12(rsin⁡θ)−36=0\Rightarrow {{\left( r\cos \theta \right)}^{2}}-12\left( r\sin \theta \right)-36=0
The above transformed equation upon simplification gives,
⇒r2cos⁡2θ−12rsin⁡θ−36=0\Rightarrow {{r}^{2}}{{\cos }^{2}}\theta -12r\sin \theta -36=0
The above equation now becomes a quadratic equation in r. Thus, we need to use the Sridhar Acharya formula which is x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . Replacing x with r, a with cos⁡2θ{{\cos }^{2}}\theta , b with −12sin⁡θ-12\sin \theta and c with −36-36 , we get,
⇒r=−(−12sin⁡θ)±(−12sin⁡θ)2−4(cos⁡2θ)(−36)2cos⁡2θ ⇒r=12sin⁡θ±144sin⁡2θ+144cos⁡2θ2cos⁡2θ ⇒r=12sin⁡θ±12sin⁡2θ+cos⁡2θ2cos⁡2θ ⇒r=12sin⁡θ±122cos⁡2θ ⇒r=6sin⁡θ±6cos⁡2θ \begin{aligned} & \Rightarrow r=\dfrac{-\left( -12\sin \theta \right)\pm \sqrt{{{\left( -12\sin \theta \right)}^{2}}-4\left( {{\cos }^{2}}\theta \right)\left( -36 \right)}}{2{{\cos }^{2}}\theta } \\\ & \Rightarrow r=\dfrac{12\sin \theta \pm \sqrt{144{{\sin }^{2}}\theta +144{{\cos }^{2}}\theta }}{2{{\cos }^{2}}\theta } \\\ & \Rightarrow r=\dfrac{12\sin \theta \pm 12\sqrt{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }}{2{{\cos }^{2}}\theta } \\\ & \Rightarrow r=\dfrac{12\sin \theta \pm 12}{2{{\cos }^{2}}\theta } \\\ & \Rightarrow r=\dfrac{6\sin \theta \pm 6}{{{\cos }^{2}}\theta } \\\ \end{aligned}
Therefore, we can conclude that the polar form of the given equation is r=6sin⁡θ±6cos⁡2θr=\dfrac{6\sin \theta \pm 6}{{{\cos }^{2}}\theta }.

Note: For conversions of this type, we need to remember the basic conversion formulae, otherwise we won’t be able to solve them within a short time. During conversion, we should take care of the signs, the degrees of the terms and the various trigonometric identities. We should also remember to simplify the converted equation as far as possible or our answer would remain incomplete, like in this problem, we applied the Sridhar Acharya formula in the converted equation in order to simplify it.