The normal at left( 2, frac{3}{2} right) to t | vedclass

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(A) The given ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$. Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 3$

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$y^2 = -104x$

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(A) The given ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$. Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 3$. 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 $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $(x_1, y_1) = \left( 2, \frac{3}{2} \right)$,$a^2 = 16$,and $b^2 = 3$: $\frac{16x}{2} - \frac{3y}{3/2} = 16 - 3$ $8x - 2y = 13$ $2y = 8x - 13 \Rightarrow y = 4x - \frac{13}{2}$. $A$ line $y = mx + c$ touches the parabola $y^2 = 4Ax$ if $c = \frac{A}{m}$. Here $m = 4$ and $c = -\frac{13}{2}$. $-\frac{13}{2} = \frac{A}{4} \Rightarrow A = -26$. Thus,the equation of the parabola is $y^2 = 4(-26)x = -104x$.

The normal at $\left( 2, \frac{3}{2} \right)$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ touches a parabola,whose equation is

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