The normal drawn at the point left(sqrt{9} co | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$. The point given is $(x_1, y_1) = \left(3 \cos \f

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$\left(\sqrt{\frac{2}{9}}, 0\right)$

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(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$. The point given is $(x_1, y_1) = \left(3 \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right) = \left(\frac{3}{\sqrt{2}}, \sqrt{\frac{7}{2}}\right)$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Substituting the values: $\frac{9x}{3/\sqrt{2}} - \frac{7y}{\sqrt{7/2}} = 9 - 7$. This simplifies to $3\sqrt{2}x - \sqrt{14}y = 2$. To find the intersection with the major axis ($x$-axis),set $y = 0$: $3\sqrt{2}x = 2 \implies x = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$. Thus,the point of intersection is $\left(\sqrt{\frac{2}{9}}, 0\right)$.

The normal drawn at the point $\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{7} = 1$ intersects its major axis at the point

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