How to Solve Hyperbola Questions Using Parametric Equations
Hyperbola is a subdivision of conic sections in the field of Mathematics. When the surface of a cone intersects a plane, curves are formed, and these curves are known as conic sections. There are three categories of conic sections: the eclipse, the hyperbola, and the parabola.
We use conic sections to study 3D geometry which has a vast number of applications in various fields of engineering. Also, when a spacecraft uses the gravitational slingshot technique, the path followed by the craft is a hyperbola. In this article, we will get to know about the parametric form of the hyperbola. We will also see some interesting facts about the hyperbola and also answer some of the questions.
Parametric Coordinates of Hyperbola
In this topic, we will find how to calculate the parametric equation of the hyperbola.
To understand this in a simple way, let’s take the help of the following diagram.
Parametric Coordinates of Hyperbola
The circle inscribed in between the hyperbola and on the transverse axis is called an auxiliary circle. If the equation of the hyperbola shown in the figure is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ then the equation of the auxiliary circle inscribed between the auxiliary circle will be $x^{2}+y^{2}=a^{2}$.
Let M$(x,y)$ be any point on the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$.
Now, from M make a line MQ to the transverse axis such that it is perpendicular to it and then select a point on the circle $x^{2}+y^{2}=a^{2}$ such that the angle ∠OCQ is 90 degrees.
Now, join (C and Q) and (O and C), The length of OC is a. Now let the length ∠SOC,
Here the ∠SOC is called the eccentric angle of the point M on the hyperbola.
Now from the right triangle OCQ, we get
$\dfrac{OC}{OQ}=cos\Theta$
$\dfrac{a}{OQ}=\dfrac{1}{sec\Theta }$ , (The radius of the circle is a=OC)
$OQ=asec\Theta$
Therefore, the abscissa of the M(x,y) on the hyperbola is asecθ.
As the point M lies on the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, so it will satisfy the equation .
So, $\dfrac{(asec\Theta )^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ (as x = sec)
$\Rightarrow \dfrac{y^{2}}{b^{2}}=sec^{2}\Theta -1$
$\Rightarrow \dfrac{y^{2}}{b^{2}}=tan^{2}\Theta $
$\Rightarrow y^{2}=b^{2}tan^{2}\Theta $
$\Rightarrow y=btan\theta$
Hence, the coordinate of the point M is (asecθ,btanθ) and for all the values of θ This point lies on the hyperbola and hence the polar coordinates of a hyperbola is represented by $(asec\Theta ,btan\Theta )$.
Parametric Form of Hyperbola
If we want to write a parametric form of the hyperbola, we can write it as
Case 1: if the hyperbola is horizontal then:
$f(t)=(x(t),y(t))$
$x(t)=asec\Theta $
$y(t)=btan\Theta$
Case 2: if the hyperbola is vertical then:
$f(t)=(x(t),y(t))$
$x(t)=atan\Theta $
$y(t)=asec\Theta $
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. When 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
Q1. Find the parametric coordinates of the point $(3\sqrt{2},2)$ on the hyperbola $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$.
Ans The equation of the hyperbola is $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$.
If we compare it with general equation of hyperbola
$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ we get
$a^{2}=9\Rightarrow a=3$
$b^{2}=4\Rightarrow b=2$
So the parametric coordinates of the hyperbola will be $(3sec\Theta ,2tan\Theta )$.
Now, to calculate the parametric coordinates,
$3sec\Theta =3\sqrt{2}\Rightarrow sec\Theta =\sqrt{2}\Rightarrow \Theta =45^{\circ}$
$2tan\Theta =2\Rightarrow tan\Theta =1\Rightarrow \Theta =45^{\circ}$
So the parametric coordinates will be $(3sec45^{\circ},2tan45^{\circ})$.
Practice Questions
Question Write parametric for each of the hyperbolas below:
1. $\dfrac{(x-3)^{2}}{256}-\dfrac{(y+3)^{2}}{625}=1$
Ans:
$x=3+16sect$
$y=-5+25tant$
2. $\dfrac{(y+7)^{2}}{324}-\dfrac{(x-8)^{2}}{169}=1$
Ans:
$x=8+13sect$
$y=-7+18tant$
Summary
The article summarises the concept of parametric coordinates of hyperbola and its form.
Now we know that if the equation a hyperbola is
$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ then the parametric coordinates will be
$(x(t)=asec\Theta,y(t)=btan\Theta)$ where $\theta$ is the parameter.
FAQs on Parametric Form of Hyperbola Explained
1. What are the parametric equations for a standard hyperbola?
The parametric equations represent the coordinates (x, y) of any point on a hyperbola in terms of a single variable, called a parameter. For a standard horizontal hyperbola with the equation x²⁄a² - y²⁄b² = 1, the parametric equations are:
- x = a secθ
- y = b tanθ
Here, 'θ' is the parameter, known as the eccentric angle. For every value of θ, the point (a secθ, b tanθ) lies on the hyperbola.
2. How are the parametric coordinates of a hyperbola derived?
The derivation involves using an auxiliary circle, which is a circle with its diameter as the transverse axis of the hyperbola. For a hyperbola x²⁄a² - y²⁄b² = 1, its auxiliary circle is x² + y² = a². By using trigonometric relationships involving a point on the hyperbola and a corresponding point on the auxiliary circle, we can express the x-coordinate as x = a secθ. Substituting this into the hyperbola's equation, we can solve for y to get y = b tanθ. This method confirms that these coordinates satisfy the hyperbola's equation for all values of θ, as sec²θ - tan²θ = 1.
3. What is the parametric form for a rectangular hyperbola like xy = c²?
A rectangular hyperbola is a special case where the asymptotes are perpendicular. For the standard form xy = c², the parametric equations are particularly simple:
- x = ct
- y = c/t
Here, 't' is the parameter. You can verify this by multiplying the coordinates: x * y = (ct) * (c/t) = c², which satisfies the original equation for any non-zero value of 't'.
4. Why is the parametric form of a hyperbola useful in mathematics and physics?
The parametric form is highly useful because it simplifies complex problems by reducing two variables (x and y) to a single parameter (θ or t). This is important for:
- Calculus: It makes finding derivatives, tangents, normals, and calculating arc lengths or areas much more manageable.
- Physics and Astronomy: It helps describe the trajectory of objects. For example, the path of a spacecraft performing a gravitational slingshot around a planet follows a hyperbolic trajectory, which can be modelled using parametric equations.
- Engineering: The shape of cooling towers is often hyperbolic (a hyperboloid of revolution) for structural strength and material efficiency, and parametric equations are used in their design.
5. How does the parametric form of a hyperbola differ from that of an ellipse?
The difference highlights the fundamental geometric distinction between the two curves.
- An ellipse is a closed curve, and its parametric equations use bounded trigonometric functions: x = a cosθ and y = b sinθ. As θ varies, the point (x, y) traces a closed path.
- A hyperbola is an open curve with two separate branches, and its equations use unbounded functions: x = a secθ and y = b tanθ. The secant and tangent functions are not defined for all angles and approach infinity, which perfectly models the infinite branches of the hyperbola.
6. How can you find the equation of a tangent to a hyperbola using its parametric form?
Using the parametric form simplifies finding the equation of a tangent at a specific point on the hyperbola. For the hyperbola x²⁄a² - y²⁄b² = 1, the equation of the tangent at the point (a secθ, b tanθ) is given by the formula:
(x⁄a)secθ - (y⁄b)tanθ = 1
This formula provides a direct way to write the tangent's equation without needing to use calculus to find the slope first, making it a powerful tool for solving problems in coordinate geometry.
7. Can you provide an example of finding the parameter for a given point on a hyperbola?
Certainly. Consider the hyperbola x²⁄9 - y²⁄4 = 1 and the point (3√2, 2) on it. Here, a = 3 and b = 2. The parametric coordinates are (3secθ, 2tanθ).
To find the parameter θ, we set the coordinates equal:
- x-coordinate: 3secθ = 3√2 => secθ = √2
- y-coordinate: 2tanθ = 2 => tanθ = 1
Both conditions are satisfied when θ = 45° or π/4 radians. Therefore, the parameter for the point (3√2, 2) is 45°.
