Tangents to Hyperbolas - Practice Questions & MCQ

Equation of Tangent of Hyperbola in Point Form:

The equation of tangent to the hyperbola, $\frac{x^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ at point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1$

Differentiating $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ w.r.t. $x$, we have

$
\begin{array}{ll} 
& \frac{2 x}{a^2}-\frac{2 y}{b^2} \frac{d y}{d x}=0 \\
\Rightarrow & \frac{d y}{d x}=\frac{b^2 x}{a^2 y} \\
\Rightarrow \quad & \left(\frac{d y}{d x}\right)_{(x, y)}=\frac{b^2 x_1}{a^2 y_1}
\end{array}
$
Hence, equation of the tangent is $y-y_1=\frac{b^2 x_1}{a^2 y_1}\left(x-x_1\right)$
or

$
\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}
$
But $\left(x_1, y_1\right)$ lies on the hyperbola $\Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$
Hence, equation of the tangent is

$
\begin{gathered}
\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1 \\
\text { or } \quad \frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1=0 \text { or } T=0
\end{gathered}
$

where $\quad T=\frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1$

Equation of Tangent of Hyperbola in Parametric Form and Slope Form

Parametric Form

The equation of tangent to the hyperbola, $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \mathrm{at}(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$ is $\frac{\mathrm{x}}{\mathrm{a}} \sec \theta-\frac{\mathrm{y}}{\mathrm{b}} \tan \theta=1$
(This can easily be derived by putting $\mathrm{x}_1=\mathrm{a} \sec \theta$ and $\mathrm{y}_1=\mathrm{b} \tan \theta$ in the point form of tangent)

Slope Form
rbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $c^2=a^2 m^2-b^2$. So the equation of tangent is $y=m x \pm \sqrt{a^2 m^2-b^2}$.
These equations are equations of two parallel tangents to hyperbola having slope $m$.