The equation of the normal to the hyperbola f | vedclass

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Question

(A) Given the hyperbola equation: $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Differentiating with respect to $x$: $\frac{2x}{16} - \frac{2y}{9} \frac{dy}{d

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$y = 0$

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(A) Given the hyperbola equation: $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Differentiating with respect to $x$: $\frac{2x}{16} - \frac{2y}{9} \frac{dy}{dx} = 0$. Thus,$\frac{dy}{dx} = \frac{9x}{16y}$. The slope of the tangent at $(-4, 0)$ is undefined (vertical tangent). Since the tangent at $(-4, 0)$ is a vertical line $(x = -4)$,the normal at this point must be a horizontal line passing through $(-4, 0)$. $A$ horizontal line passing through $(-4, 0)$ has the equation $y = 0$.

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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