The locus of the point of intersection of mut | vedclass

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Any tangent to the ellipse is given by $y = mx \pm \sqrt{a^2m^2 + b^2}

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$A$ circle

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Any tangent to the ellipse is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$. If this tangent passes through a point $(h, k)$,then $k - mh = \pm \sqrt{a^2m^2 + b^2}$. Squaring both sides,we get $(k - mh)^2 = a^2m^2 + b^2$,which simplifies to $k^2 - 2mhk + m^2h^2 = a^2m^2 + b^2$. Rearranging as a quadratic in $m$: $m^2(h^2 - a^2) - 2mhk + (k^2 - b^2) = 0$. Since the tangents are mutually perpendicular,the product of their slopes $m_1m_2 = -1$. For the quadratic equation $Am^2 + Bm + C = 0$,the product of roots is $\frac{C}{A}$. Thus,$\frac{k^2 - b^2}{h^2 - a^2} = -1$. $k^2 - b^2 = -(h^2 - a^2) \implies h^2 + k^2 = a^2 + b^2$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = a^2 + b^2$,which represents a circle known as the director circle.

The locus of the point of intersection of mutually perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

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