The locus of the foot of the perpendicular dr | vedclass

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(A) The given equation of the ellipse is $x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.Here,$a^2 = 6$ and $b^2 = 2$.The

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$(x^2 + y^2)^2 = 6x^2 + 2y^2$

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(A) The given equation of the ellipse is $x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.Here,$a^2 = 6$ and $b^2 = 2$.The equation of any tangent to the ellipse is $y = mx + \sqrt{a^2m^2 + b^2}$,which becomes $y = mx + \sqrt{6m^2 + 2}$  $(1)$.The line perpendicular to this tangent passing through the origin $(0, 0)$ has the equation $y = -\frac{1}{m}x$,which implies $m = -\frac{x}{y}$  $(2)$.Substituting the value of $m$ from $(2)$ into $(1)$:$y = (-\frac{x}{y})x + \sqrt{6(-\frac{x}{y})^2 + 2}$$y + \frac{x^2}{y} = \sqrt{\frac{6x^2 + 2y^2}{y^2}}$$\frac{y^2 + x^2}{y} = \frac{\sqrt{6x^2 + 2y^2}}{|y|}$Squaring both sides,we get $(x^2 + y^2)^2 = 6x^2 + 2y^2$.

The locus of the foot of the perpendicular drawn from the centre of the ellipse $x^2 + 3y^2 = 6$ to any tangent to it is:

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