If a normal drawn to the ellipse frac{x^2}{4} | vedclass

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(A) 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 $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1}

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$3$

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(A) 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 $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Here $a^2 = 4$ and $b^2 = 3$,so the normal is $\frac{4x}{x_1} - \frac{3y}{y_1} = 1$. The slope of this normal is $m = \frac{4y_1}{3x_1}$. This line is a tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$. The condition for the line $y = mx + c$ to be a tangent to $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ is $c^2 = A^2 m^2 - B^2$. Rewriting the normal as $y = \frac{4y_1}{3x_1} x - \frac{y_1}{3}$,we have $m = \frac{4y_1}{3x_1}$ and $c = -\frac{y_1}{3}$. Substituting into the condition: $(-\frac{y_1}{3})^2 = 4(\frac{4y_1}{3x_1})^2 - 3$. Since $(x_1, y_1)$ lies on the ellipse,$\frac{x_1^2}{4} + \frac{y_1^2}{3} = 1$,so $x_1^2 = 4(1 - \frac{y_1^2}{3}) = \frac{4(3 - y_1^2)}{3}$. Substituting $x_1^2$ into the condition leads to $m^2 = 3$.

If a normal drawn to the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ touches the hyperbola $\frac{x^2}{4} - \frac{y^2}{3} = 1$,then the square of the slope of that normal is

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