Question

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)

• A. $be$
• B. $e$
• C. $ab$
• D. $ae$

Answer 1

The correct option is B $e$

Let a hyperbola , $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$---------1

Let the Eccentricity be e.

End point or extremities of LR are $(ae,\frac{\pm b^2}{a})$

Tangent at L$(ae,\frac{b^2}{a})$ will be

$\Rightarrow \frac{x\left(ae\right)}{a^2}-\frac{y\left(\frac{b^2}{a}\right)}{b^2}=1$ (as tangent at $(x_1,y_1)$ is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$)

$\Rightarrow \frac{xe}{a}-\frac{y}{a}=1$

$\Rightarrow xe-y=a\Rightarrow y=xe-a$--------2

General equation of a tangent in slope form $y=mx+c$------3

Comparing Equation 2 & 3 , as they represent the tangents of Hyperbola,

$m=e$

The gradient i.e. slope of tangent$=m=e$