The magnitude of the gradient of the tangent | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the extremities of the latera recta are $(ae, \pm \fr

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$e$

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the extremities of the latera recta are $(ae, \pm \frac{b^2}{a})$. Let us consider the point $P(ae, \frac{b^2}{a})$. The equation of the tangent at $(x_1, y_1)$ to the hyperbola is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Substituting $(x_1, y_1) = (ae, \frac{b^2}{a})$,we get: $\frac{x(ae)}{a^2} - \frac{y(b^2/a)}{b^2} = 1$ $\Rightarrow \frac{xe}{a} - \frac{y}{a} = 1$ $\Rightarrow xe - y = a$ $\Rightarrow y = ex - a$. Comparing this with the slope-intercept form $y = mx + c$,the slope $m$ is $e$. Thus,the magnitude of the gradient is $|e| = e$.

The magnitude of the gradient of the tangent at an extremity of the latera recta of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola).

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