(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{6} = 1$,where $a^2 = 4$ and $b^2 = 6$. The locus of the intersection of perpendicular tangents to

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(A) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{6} = 1$,where $a^2 = 4$ and $b^2 = 6$. The locus of the intersection of perpendicular tangents to an ellipse is its director circle,given by $x^2 + y^2 = a^2 + b^2$. Here,$a^2 = 4$ and $b^2 = 6$. Therefore,the equation of the director circle is $x^2 + y^2 = 4 + 6 = 10$. Since $(\alpha, \beta)$ is the point of intersection of two perpendicular tangents,it must lie on the director circle. Thus,$\alpha^2 + \beta^2 = 10$.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Ellipse

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Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

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$26$
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$5/12$

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10-2. Parabola, Ellipse, Hyperbola

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Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If $(\alpha, \beta)$ is the point where $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly,then $\alpha^2+\beta^2=$

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