If a tangent having slope m = frac{1}{3} to t | vedclass

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(A) The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given $m = \fr

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$\left(\frac{2}{5}, 4\right)$

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(A) The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Given $m = \frac{1}{3}$,the tangent is $y = \frac{1}{3}x \pm \sqrt{\frac{a^2}{9} + b^2}$,which simplifies to $x - 3y \pm 3\sqrt{\frac{a^2}{9} + b^2} = 0$.This line is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,which has center $(-1, -1)$ and radius $r = 1$.$A$ line is a normal to a circle if it passes through the center of the circle.Substituting $(-1, -1)$ into the tangent equation: $-1 - 3(-1) \pm 3\sqrt{\frac{a^2}{9} + b^2} = 0$,so $2 \pm 3\sqrt{\frac{a^2}{9} + b^2} = 0$.This implies $3\sqrt{\frac{a^2}{9} + b^2} = 2$,so $\frac{a^2}{9} + b^2 = \frac{4}{9}$,or $a^2 + 9b^2 = 4$.Since $a > b > 0$,we have $b^2 = \frac{4 - a^2}{9}$.Since $b^2 > 0$,$4 - a^2 > 0 \implies a^2 Since $a^2 > b^2$,$a^2 > \frac{4 - a^2}{9} \implies 9a^2 > 4 - a^2 \implies 10a^2 > 4 \implies a^2 > \frac{2}{5}$.Thus,$a^2 \in \left(\frac{2}{5}, 4\right)$.

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

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