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Tangents to Hyperbolas - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Equation of Tangent of Hyperbola in Point Form, Equation of Tangent of Hyperbola in Parametric Form and Slope Form is considered one of the most asked concept.
63 Questions around this concept.

Solve by difficulty

The tangent at a point P on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$ meets one of the directrices in F. If PF subtends
an angle $\mathrm{\theta }$ at the corresponding focus, then $\mathrm{\theta }$ equals

$\mathrm{\pi / 4}$

$\mathrm{\pi / 2}$

$\mathrm{3\pi / 4}$

$\mathrm{\pi }$

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Tangents are drawn to $\mathrm{3 x^2-2 y^2=6}$ from a point P. If these tangents intersect the coordinate axes at concyclic points, The locus of P is

$\mathrm{\frac{1}{x^2}+\frac{1}{y^2}=5}$

$\mathrm{x^2-y^2=5}$

$\mathrm{\frac{1}{x^2}-\frac{1}{y^2}=5}$

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Concepts Covered - 0

Equation of Tangent of Hyperbola in Point Form

Equation of Tangent of Hyperbola in Point Form:

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

$\\ {\text { Differentiating } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { w.r.t. } x, \text { we have }} \\\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0}\\\\ {\Rightarrow \quad \;\;\;\;\;\;\;\;\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}} \\\\ {\Rightarrow \quad\;\;\;\;\;\;\;\left(\frac{d y}{d x}\right)_{(x, y)}=\frac{b^{2} x_{1}}{a^{2} y_{1}}} \\\\ \text { Hence, equation of the tangent is } y-y_{1}=\frac{b^{2} x_{1}}{a^{2} y_{1}}\left(x-x_{1}\right) \\\\ \mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}}\\\\ \text { But }\left(x_{1}, y_{1}\right) \text { lies on the hyperbola } \Rightarrow \frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1$

$\\ {\text { Hence, equation of the tangent is }} \\\\ {\;\;\;\;\;\;\;\;\;\;\;\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1}\\\\ \text { or } \quad \frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}-1=0 \text { or } T=0\\\\ \text { where } \quad T=\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}-1$

Note:

T = 0 can be used to get the equation of tangent on the point (x₁, y₁) lying on any general hyperbola as well.

Equation of Tangent of Hyperbola in Parametric Form and Slope Form

Equation of Tangent of Hyperbola in Parametric Form and Slope Form

Parametric Form
$\\ {\text {The equation of tangent to the hyperbola, } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \text { at }(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta) \text { is }} \\ {\frac{\mathrm{x}}{\mathrm{a}} \sec \theta-\frac{\mathrm{y}}{\mathrm{b}} \tan \theta=1}$

(This can easily be derived by putting x₁ = a sec $\theta$ and y₁ = b tan $\theta$ in the point form of tangent)

Slope Form

We have studied that if the line y = mx + c is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , then c² = a²m² - b². So the equation of tangent is $y=mx\pm\sqrt{a^2m^2-b^2}$ .