Various Forms of Tangents in Hyperbola

A hyperbola is a set of all points (x, y) such that the difference of the distances between (x, y) and two different points is constant. A hyperbola has two vertices that lie on the axis of symmetry known as the transverse axis. The transverse axis of the hyperbola can either be horizontal or vertical. In this article, we will get to know about the different types of equations of the tangent to hyperbola like the equation of tangent of hyperbola in slope form, equation of tangent of hyperbola in parametric form, the chord of contact of hyperbola, and point of contact of the tangent to a hyperbola.

Equation of Hyperbola

The general equation of hyperbola can be represented as $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$.

Hyperbola

Point Form Equation of a Tangent to Hyperbola

In this form, the tangent is drawn from the point of contact of the tangent to the hyperbola. Let’s say the point of contact of the hyperbola to tangent is $(x_{1},y_{1})$, then the equation of the tangent to hyperbola will be $\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}=1$.

Equation of Tangent to Hyperbola in Slope Form

This type of equation gives us the equation of the tangent of hyperbola in terms of the slope of the line “m”. The equation is $y=mx\pm \sqrt{a^{2}m^{2}-b^{2}}$

This equation is also called “The condition of tangency”.

Equation of Pair of Tangents in Hyperbola

When the equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, then the pair of tangents can be represented using $SS_{1}=T^{2}$ i.e

$(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}-1)(\dfrac{x_{1}^{2}}{a^{2}}-\dfrac{y_{1}^{2}}{b^{2}}-1)=(\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}-1)$

Chord of Contact of Hyperbola

A chord of contact is a chord passing through endpoints of tangents drawn from a point

$(x_{1},y_{1})$ to the hyperbola. The equation of chord of contact of hyperbola will be $\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}=1$ .

Equation of Tangent to Hyperbola in Parametric Form

The parametric coordinates of any hyperbola can be represented as$a sec\theta, btan\theta$ and the equation of a tangent to hyperbola will be $\dfrac{x(asec\Theta )}{a^{2}}-\dfrac{y(atan\Theta )}{b^{2}}=1$

Interesting Facts

• When an object, let's say a jet, moves faster than the speed of sound, it creates a conical form of a wave in space and that wave intersects the ground, the curve we get from that intersection is a hyperbola.

• The cooling towers are generally made of hyperbolic shape to achieve 2 things: first, the least amount of material used to make it and second, the structure should be strong enough to withstand strong winds.

Solved Examples

Example 1. Find the equation of a tangent to the hyperbola $x^{2}-4y^{2}=4$ which is parallel to the line $x+2y=0$.

Solution: Equation of hyperbola : $\dfrac{x^{2}}{4}-\dfrac{y^{2}}{1}=1$ ,

So ,$a^{2}=4\Rightarrow a=2$

$b^{2}=1\Rightarrow a=1$

The slope of the given line will be $\dfrac{1}{2}$. Now, using the condition of tangency we will calculate the value of c.

$c^{2}=a^{2}m^{2}-b^{2}$

$c^{2}=2^{2}(-\dfrac{1}{2}^{2})-(1)^{2}$

$c=0$

So, the equation of tangent will be

$y=-\dfrac{1}{2}x$

$x+2y=0$

Example 2. Find the equation of the chord of contact of the hyperbola $\dfrac{x^{2}}{6}-\dfrac{y^{2}}{2}=1$ if the tangents are drawn from (3,2).

Solution: We know that The equation of chord of contact of a hyperbola is $\dfrac{xx_{1}}{a^{2}}-\dfrac{yy_{1}}{b^{2}}=1$. So,

$\dfrac{x(3)}{6}-\dfrac{y(2)}{2}=1$

$\dfrac{x}{2}-\dfrac{y}{1}=1$

$x-2y=2$is the required equation of chord of contact.

Practice Questions

Question 1. What is the value of m for which $y=mx+6$ is tangent to the hyperbola $\dfrac{x^{2}}{100}-\dfrac{y^{2}}{49}=1$?

Ans: $\sqrt{\dfrac{17}{20}}$,$- \sqrt{\dfrac{17}{20}}$

Question 2. A common tangent to $9x^{2}-16y^{2}=144$ and $x^{2}+y^{2}=9$ is____.

Ans: $y=3\sqrt{\dfrac{2}{7}}x\pm \dfrac{15}{\sqrt{7}}$

Summary

The article summarises the concept of the chord of contact and tangents as a hyperbola. We learnt about different types of forms of tangents and how to find the equation of these tangents, then we did some examples to brush up on our concepts and get a better understanding of the topic. We hope to have helped you clear your doubts on this topic and learn something new. Do try out the solved examples and practise questions to evaluate your understanding of the concepts discussed here.