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 $({\mathrm{x}}_1,{\mathrm{y}}_1)$ 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

$\begin{array}{ll}& \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0\\ \Rightarrow & \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}\\ \Rightarrow & {\left(\frac{dy}{dx}\right)}_{(x,y)}=\frac{b^2{x}_1}{a^2{y}_1}\end{array}$
Hence, equation of the tangent is $y-y_1=\frac{b^2{x}_1}{a^2{y}_1}(x-x_1)$
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 $(x_1,y_1)$ lies on the hyperbola $\Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$
Hence, equation of the tangent is

$\begin{array}{c}\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1\\ \text{ or }\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-1=0\text{ or }T=0\end{array}$

where $T=\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-1$

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 $y=mx\pm \sqrt{a^2{m}^2-b^2}$.
These equations are equations of two parallel tangents to hyperbola having slope $m$.