The locus of the feet of the perpendiculars f | vedclass

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(C) The given equation of the ellipse is $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$. Dividing by $20$,we get $\frac{(x - y + 1)^2}{20} + \frac{(2x + 2y -

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$x^2 + y^2 - 2x - 4y - 5 = 0$

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(C) The given equation of the ellipse is $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$. Dividing by $20$,we get $\frac{(x - y + 1)^2}{20} + \frac{(2x + 2y - 6)^2}{20} = 1$. This can be rewritten as $\frac{(\frac{x - y + 1}{\sqrt{2}})^2}{10} + \frac{(\frac{x + y - 3}{\sqrt{2}})^2}{5} = 1$. Here,$a^2 = 10$ and $b^2 = 5$. The center of the ellipse is $(1, 2)$. The locus of the feet of the perpendiculars from the foci to any tangent of an ellipse is its auxiliary circle. The auxiliary circle has the same center as the ellipse and radius equal to the semi-major axis $a$. Here,$a^2 = 10$,so the equation of the auxiliary circle is $(x - 1)^2 + (y - 2)^2 = 10$. Expanding this,we get $x^2 - 2x + 1 + y^2 - 4y + 4 = 10$,which simplifies to $x^2 + y^2 - 2x - 4y - 5 = 0$.

The locus of the feet of the perpendiculars from either focus of the ellipse $(x - y + 1)^2 + (2x + 2y - 6)^2 = 20$ onto any tangent is:

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