Calculus 10th Edition

a.) To find the equation of the hyperbola we need a point and the slope of the curve at that point. We are told that the point (3, -2) but we do not have the point so we need to differentiate the hyperbola function and evaluate it at (3, -2) to find the slope at (3, -2). 2($\dfrac{x}{6}$) - 2($\dfrac{y}{8})(\dfrac{dy}{dx}$) = 0 $\dfrac{x}{3}$ - $(\dfrac{y}{4})(\dfrac{dy}{dx})$ = 0 $\dfrac{x}{3}$ = $(\dfrac{y}{4})(\dfrac{dy}{dx})$ $\dfrac{dy}{dx}$ = $\dfrac{\dfrac{x}{3}}{\dfrac{y}{4}}$ $\dfrac{dy}{dx}$ = $\dfrac{4x}{3y}$ Evaluate the point at (3, -2) $\dfrac{dy}{dx}|_{(x, y) = (3, -2)} = \dfrac{4(3)}{3(-2)} = \dfrac{12}{-6} = -2$ Now plug in the values to point slope form, $y - y_1 = m(x - x_1)$ where m is the slope. y - (-2) = -2(x - (3)) y = -2x + 4 b.) Again, to find the equation of a tangent line we need a point and a slope. It can be inferred that our point is $(x_0, y_0)$ Now we need to find the slope at this point by implicitly differentiating the hyperbola. 2($\dfrac{x}{a^2}$) - 2($\dfrac{y}{b^2})(\dfrac{dy}{dx}$) = 0 $\dfrac{2x}{a^2}$ - $(\dfrac{2y}{b^2})(\dfrac{dy}{dx})$ = 0 $\dfrac{2x}{a^2}$ = $(\dfrac{2y}{b^2})(\dfrac{dy}{dx})$ $\dfrac{dy}{dx}$ = $\dfrac{\dfrac{2x}{a^2}}{\dfrac{2y}{b^2}}$ $\dfrac{dy}{dx}$ = $\dfrac{2xb^2}{2ya^2}$ = $\dfrac{xb^2}{ya^2}$ This is the slope at the point $(x_0, y_0)$. Now we need to plug it into the point slope equation. $(y - (y_0)) = \dfrac{xb^2}{ya^2}(x - (x_0))$ $ya^2(y - y_0) = xb^2(x - x_0)$ $\dfrac{y(y - y_0)}{b^2} = \dfrac{x(x - x_0)}{a^2}$ $\dfrac{y^2 - y_0y}{b^2} = \dfrac{x^2 - x_0x}{a^2}$ $\dfrac{y^2}{b^2} - \dfrac{y_0y}{b^2} = \dfrac{x^2}{a^2} - \dfrac{x_0x}{a^2}$ $\dfrac{x_0x}{a^2} - \dfrac{y_0y}{b^2} = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}$ The question says that $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, so $\dfrac{x_0x}{a^2} - \dfrac{y_0y}{b^2} = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\dfrac{x_0x}{a^2} - \dfrac{y_0y}{b^2} = 1$