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The pole of the straight line $x+4 y=4$ with respect to the ellipse $x^2+4 y^2=4$ is
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Verified Answer
The correct answer is:
$(1,1)$
Equation of line is $x+4 y-4=0$
$$
l=1, m=4, n=-4
$$
and equation of ellipse is
$$
\begin{aligned}
\Rightarrow & x^2+4 y^2=4 \\
\Rightarrow & \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& a^2=4, b^2=1
\end{aligned}
$$
Pole of the straight line with respect to ellipse
$$
=\left(\frac{-a^2 l}{n}, \frac{-b^2 m}{n}\right)=\left(\frac{-4.1}{-4}, \frac{-1.4}{-4}\right)=(1,1)
$$
$$
l=1, m=4, n=-4
$$
and equation of ellipse is
$$
\begin{aligned}
\Rightarrow & x^2+4 y^2=4 \\
\Rightarrow & \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& a^2=4, b^2=1
\end{aligned}
$$
Pole of the straight line with respect to ellipse
$$
=\left(\frac{-a^2 l}{n}, \frac{-b^2 m}{n}\right)=\left(\frac{-4.1}{-4}, \frac{-1.4}{-4}\right)=(1,1)
$$
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