The equation of the normal to the hyperbola f | vedclass

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(D) The equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$. Comparing this with $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we

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$y = 0$

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(D) The equation of the hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$. Comparing this with $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = 16$ and $b^{2} = 9$. The point $(-4, 0)$ lies on the hyperbola since $\frac{(-4)^{2}}{16} - \frac{0^{2}}{9} = 1 - 0 = 1$. The slope of the tangent at any point $(x_{1}, y_{1})$ is given by differentiating the hyperbola equation: $\frac{2x}{16} - \frac{2y y'}{9} = 0 \Rightarrow y' = \frac{9x}{16y}$. At $(-4, 0)$,the tangent is vertical $(y' 	o \infty)$,which means the tangent line is $x = -4$. $A$ line perpendicular to a vertical line is a horizontal line. Since the normal passes through $(-4, 0)$ and is perpendicular to the vertical tangent $x = -4$,the normal must be the horizontal line passing through $y = 0$. Thus,the equation of the normal is $y = 0$.

The equation of the normal to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at $(-4, 0)$ is

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