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If the tangents drawn from a point $P$ to the ellipse $4 x^2+9 y^2-24 x+36 y=0$ are perpendicular, then the locus of $P$ is
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Verified Answer
The correct answer is:
$x^2+y^2-6 x+4 y-13=0$
Given, equation of ellipse is
$$
\begin{aligned}
& 4 x^2+9 y^2-24 x+36 y=0 \\
& \Rightarrow \quad 4\left(x^2-6 x+9\right)+9\left(y^2+4 y+4\right)-36-36=0 \\
& \Rightarrow \quad 4(x-3)^2+9(y+2)^2=72 \\
& \Rightarrow \quad \frac{(x-3)^2}{18}+\frac{(y+2)^2}{8}=1
\end{aligned}
$$
Here, required locus is director circle.
$$
\begin{array}{lrr}
\therefore & (x-3)^2+(y+2)^2=26 \\
\Rightarrow & x^2-6 x+9+y^2+4+4 y=26 \\
\Rightarrow & x^2+y^2-6 x+4 y=13 \\
\Rightarrow & x^2+y^2-6 x+4 y-13=0
\end{array}
$$
$$
\begin{aligned}
& 4 x^2+9 y^2-24 x+36 y=0 \\
& \Rightarrow \quad 4\left(x^2-6 x+9\right)+9\left(y^2+4 y+4\right)-36-36=0 \\
& \Rightarrow \quad 4(x-3)^2+9(y+2)^2=72 \\
& \Rightarrow \quad \frac{(x-3)^2}{18}+\frac{(y+2)^2}{8}=1
\end{aligned}
$$
Here, required locus is director circle.
$$
\begin{array}{lrr}
\therefore & (x-3)^2+(y+2)^2=26 \\
\Rightarrow & x^2-6 x+9+y^2+4+4 y=26 \\
\Rightarrow & x^2+y^2-6 x+4 y=13 \\
\Rightarrow & x^2+y^2-6 x+4 y-13=0
\end{array}
$$
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