If the normal to the ellipse frac{x^2}{16} + | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}

Vedclass · Accepted Answer

$y^2 = -104x$

Vedclass · Answer

(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Here,$a^2 = 16$,$b^2 = 3$,$x_1 = 2$,and $y_1 = \frac{3}{2}$. Substituting these values: $\frac{16x}{2} - \frac{3y}{3/2} = 16 - 3$. $8x - 2y = 13$,which simplifies to $2y = 8x - 13$ or $y = 4x - \frac{13}{2}$. For a line $y = mx + c$ to be tangent to a parabola $y^2 = 4Ax$,the condition is $c = \frac{A}{m}$. Here,$m = 4$ and $c = -\frac{13}{2}$. So,$-\frac{13}{2} = \frac{A}{4} \implies A = -26$. The equation of the parabola is $y^2 = 4Ax = 4(-26)x = -104x$.

If the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ at the point $\left( 2, \frac{3}{2} \right)$ is tangent to a parabola,find the equation of the parabola.

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