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(B) The given equation of the ellipse is $4 x^2+9 y^2-24 x+36 y=0$. Completing the square,we get: $4(x^2-6 x+9)+9(y^2+4 y+4) = 36+36$ $4(x-3)^2+9(y+2)

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$x^2+y^2-6 x+4 y-13=0$

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(B) The given equation of the ellipse is $4 x^2+9 y^2-24 x+36 y=0$. Completing the square,we get: $4(x^2-6 x+9)+9(y^2+4 y+4) = 36+36$ $4(x-3)^2+9(y+2)^2 = 72$ Dividing by $72$,we get: $\frac{(x-3)^2}{18}+\frac{(y+2)^2}{8}=1$. Here,$a^2=18$ and $b^2=8$. The locus of a point from which perpendicular tangents are drawn to an ellipse is its director circle,given by $(x-h)^2+(y-k)^2 = a^2+b^2$. Substituting the values,we get: $(x-3)^2+(y+2)^2 = 18+8$ $(x-3)^2+(y+2)^2 = 26$ $x^2-6 x+9+y^2+4 y+4 = 26$ $x^2+y^2-6 x+4 y+13 = 26$ $x^2+y^2-6 x+4 y-13 = 0$.

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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