The magnitude of the gradient of the tangent | vedclass

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Question

Let a hyperbola, $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1-\ldots$

Let the Eccentricity be e. End point or extremities of LR are $\left(\right.$ae

Vedclass · Accepted Answer

$e$

Vedclass · Answer

Let a hyperbola, $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1-\ldots$

Let the Eccentricity be e. End point or extremities of LR are $\left(\right.$ae, $\left.\frac{\pm b^{2}}{a}\right)$ Tangent at $L \left(\right.$ ae $\left., \frac{b^{2}}{a}\right)$ will be

$\Rightarrow \frac{x(a e)}{a^{2}}-\frac{y\left(\frac{b^{2}}{a}\right)}{b^{2}}=1\left(\right.$ as tangent at $\left(x_{1}, y_{1}\right)$ is $\left.\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1\right)$

$\Rightarrow \frac{x e}{a}-\frac{y}{a}=1$

$\Rightarrow x e-y=a \Rightarrow y=x e-a-x-2$

General equation of a tangent in slope form $y=m x+c$ -

Comparing Equation $2 \& 3$, as they represent the tangents of Hyperbola,

$m=e$

The gradient i.e. slope of tangent $=m=e$

The magnitude of the gradient of the tangent at an extremity of 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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