Question

Find the pole of the straight line 2x-y=1 with respect to the circle $5{{x}^{2}}+5{{y}^{2}}=9$.

Answer 1

Hint: Here we have been given the equation of a line and the equation of a circle with respect to which we need to find the line’s pole. We will do this by first assuming the pole to be (h,k). Then we will find the equation of the polar line by using T=0 on the circle with this point. Then we will compare this equation to the equation of the given line as these two are the same line. Hence, we will obtain the value of h and k by comparing these. Thus, we will get the required results.

Complete step-by-step solution:
We have been given the straight line $2x-y=1$ and the circle $5{{x}^{2}}+5{{y}^{2}}=9$.
Now, we need to find the pole of the given straight line with respect to the given circle.
Let us first assume the pole of this line to be (h,k).
We know that the polar line is given by using T=0.
Thus, the equation of the polar line when the pole is (h,k) is given as:
$\begin{align}
  & T=0 \\
 & \Rightarrow 5x.h+5.y.k=9 \\
 & \Rightarrow 5hx+5ky=9 \\
\end{align}$
Now, the given polar line to us is $2x-y=1$. We know that this line and the calculated polar lines are the same.
Thus, on comparing these equations, we get:
$\begin{align}
  & 5hx+5ky=9 \\
 & 2x-y=1 \\
 & \Rightarrow \dfrac{5h}{2}=\dfrac{5k}{-1}=\dfrac{9}{1} \\
\end{align}$
On solving these, we get:
$\begin{align}
  & \dfrac{5h}{2}=\dfrac{9}{1} \\
 & \Rightarrow 5h=18 \\
 & \therefore h=\dfrac{18}{5} \\
 & \dfrac{5k}{-1}=\dfrac{9}{1} \\
 & \Rightarrow 5k=-9 \\
 & \therefore k=-\dfrac{9}{5} \\
\end{align}$
Thus, the required pole of this line is $\left( \dfrac{18}{5},-\dfrac{9}{5} \right)$.

Note: We can also find the pole by directly using the formula of the pole of a line $lx+my+n=0$ with respect to the circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ given as:
$\left( -\dfrac{{{a}^{2}}l}{n},-\dfrac{{{a}^{2}}m}{n} \right)$
Here, the given line is:
$\begin{align}
  & 2x-y=1 \\
 & \Rightarrow 2x-y-1=0 \\
\end{align}$
Thus, we can see that:
l=2
m=-1
n=-1
Here, the circle given to us is:
$\begin{align}
  & 5{{x}^{2}}+5{{y}^{2}}=9 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{9}{5} \\
\end{align}$
Thus, we can see that:
${{a}^{2}}=\dfrac{9}{5}$
Thus, we get the pole as:
 $\begin{align}
  & \left( -\dfrac{{{a}^{2}}l}{n},-\dfrac{{{a}^{2}}m}{n} \right) \\
 & \Rightarrow \left( -\left( \dfrac{9}{5} \right)\left( \dfrac{2}{-1} \right),-\left( \dfrac{9}{5} \right)\left( \dfrac{-1}{-1} \right) \right) \\
 & \therefore \left( \dfrac{18}{5},-\dfrac{9}{5} \right) \\
\end{align}$