Question

Eliminate \[\theta \] from the following equations:
 \[\begin{align}
  & 1)x=h+a\cos \theta \\
 & y=k+b\sin \theta \\
\end{align}\]
\[\begin{align}
  & 2)x\cos \theta -y\sin \theta =a \\
 & x\sin \theta +y\cos \theta =b \\
\end{align}\]

Answer 1

Hint: Eliminating a term from the equations means to solve the equations together in such a way to obtain a new equation that doesn’t contain the term to be eliminated. In this question, to eliminate \[\theta \] we will simplify the equations in each part and eliminate all terms containing \[\theta \].

Formula used:
The method used in these two parts is the elimination method.
In the elimination method, firstly, each equation is multiplied by a suitable number so that the two equations have the same leading coefficient. After that, the second equation is subtracted from first and the new equation is solved to remove the term to be eliminated.
For e.g. In these two equations, first is multiplied by \[b\] and second by \[a\] to eliminate \[x\]
\[\begin{align}
  & ax+by=c \\
 & bx+fy=d \\
\end{align}\]

Complete step by step answer:
Solving part 1) Firstly, we will eliminate the term \[\theta \] from these two equations by simplifying the equations in the following manner.
 \[\begin{align}
  & x=h+a\cos \theta \\
 & \Rightarrow x-h=a\cos \theta \\
 & \Rightarrow \dfrac{x-h}{a}=\cos \theta ---\left( 1 \right) \\
 & y=k+b\sin \theta \\
 & \Rightarrow y-k=b\sin \theta \\
 & \Rightarrow \dfrac{y-k}{b}=\sin \theta ---\left( 2 \right) \\
\end{align}\]
Squaring and adding \[\left( 1 \right)\]and \[\left( 2 \right)\]
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{x-h}{a} \right)}^{2}}+{{\left( \dfrac{y-k}{b} \right)}^{2}}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\
 & \Rightarrow {{\left( \dfrac{x-h}{a} \right)}^{2}}+{{\left( \dfrac{y-k}{b} \right)}^{2}}=1 \\
\end{align}\]
Hence, as we see in the last equation, \[\theta \] has been eliminated.
In part
\[\begin{align}
  & 2)x\cos \theta -y\sin \theta =a---\left( 3 \right) \\
 & x\sin \theta +y\cos \theta =b---\left( 4 \right) \\
\end{align}\]
First of all, multiplying equation (3) by x and (4) by y and adding
\[\begin{align}
  & {{x}^{2}}\cos \theta -xy\sin \theta =ax \\
 & xy\sin \theta +{{y}^{2}}\cos \theta =by \\
 & \Rightarrow \cos \theta \left( {{x}^{2}}+{{y}^{2}} \right)=ax+by \\
 & \Rightarrow \cos \theta =\dfrac{ax+by}{{{x}^{2}}+{{y}^{2}}} \\
\end{align}\]
Here, we obtained the value of cosine function
Secondly, multiplying first equation by y and second by x and subtracting (3) from (4)
\[\begin{align}
  & xy\cos \theta -{{y}^{2}}\sin \theta =ay \\
 & _{-}{{x}^{2}}\sin \theta {{+}_{-}}xy\cos \theta {{=}_{-}}bx \\
 & \Rightarrow -\left( {{x}^{2}}+{{y}^{2}} \right)\sin \theta =ay-bx \\
 & \Rightarrow \sin \theta =\dfrac{bx-ay}{{{x}^{2}}+{{y}^{2}}} \\
\end{align}\]
Hence, we got the value of sine function also
Squaring and adding the sine and cosine terms
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta ={{\left( \dfrac{bx-ay}{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}+{{\left( \dfrac{ax+by}{{{x}^{2}}+{{y}^{2}}} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{bx-ay}{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}+{{\left( \dfrac{ax+by}{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}=1 \\
\end{align}\]
Thus, we have eliminated the \[\theta \] terms from the above equations.

Note: If you don’t have equations where you can eliminate a variable by addition or subtraction, you can directly begin by multiplying one or both the equations with a constant to obtain an equivalent linear system where one of the variables can be eliminated.