If the normal drawn at the point (2, -1) to t | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{8} + \frac{y^2}{2} = 1$. The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \

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$14$

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(B) The equation of the ellipse is $\frac{x^2}{8} + \frac{y^2}{2} = 1$. The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Substituting $(x_1, y_1) = (2, -1)$,$a^2 = 8$,and $b^2 = 2$: $\frac{8x}{2} - \frac{2y}{-1} = 8 - 2 $ $\Rightarrow 4x + 2y = 6 $ $\Rightarrow y = 3 - 2x$. Substituting $y = 3 - 2x$ into the ellipse equation $x^2 + 4y^2 = 8$: $x^2 + 4(3 - 2x)^2 = 8 $ $\Rightarrow x^2 + 4(9 - 12x + 4x^2) = 8 $ $\Rightarrow 17x^2 - 48x + 28 = 0$. Since $(2, -1)$ is a point on the ellipse,$x = 2$ is a root of this quadratic equation. Let the other root be $a$. Using the product of roots formula $x_1 x_2 = \frac{c}{A}$: $2 	imes a = \frac{28}{17} $ $\Rightarrow a = \frac{14}{17} $ $\Rightarrow 17a = 14$.

If the normal drawn at the point $(2, -1)$ to the ellipse $x^2 + 4y^2 = 8$ meets the ellipse again at $(a, b)$,then $17a =$

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