Ellipse Vs Hyperbola

Conic Sections is an extremely important topic of IIT JEE Mathematics. The conics like circle, parabola, ellipse and hyperbola are all interrelated and therefore it is crucial to know their distinguishing features as well as similarities in order to attempt the questions in various competitive exams like the JEE. Hence, students are advised to study these topics attentively in order to remain competitive in the JEE.

In this section, we shall discuss the similarities and differences between ellipse and hyperbola in detail.  

The standard equation of ellipse is x²/a² + y²/b² = 1, where a > b and b² = a²(1 – e²), while that of a hyperbola is x²/a² – y²/b² = 1, where b² = a²(e² -1). Hence, the equations are closely related, the only difference being of a negative sign. 

Both ellipses as well as hyperbolas have vertices, foci, and a center. In standard form, both the curves have (0, 0) as the centre. As discussed above, in an ellipse, ‘a’ is always greater than b. if ‘a’ is greater than ‘b’ and ‘a’ lies below the term of x² then the major axis is horizontal and similarly, if it lies under the y² term, then  the axis is vertical.

The situation is quite different in case of a hyperbola. In hyperbola, ‘a’ may be greater than, equal to or less than ‘b’. The order of the terms x² and y² decide whether the transverse axis would be horizontal or vertical. If x² comes first then the transverse axis would be horizontal. Conversely, if the y² term is listed first, the transverse axis would be vertical.

Going into the intricacies, we see that the focal distance in hyperbola is greater than the distance of the vertex. The situation is quite different in case of ellipse; the distance from the centre to vertex is greater than the focal distance.

Another striking difference between the both is that hyperbolas have asymptotes which ellipses don’t have. 

We now proceed towards the various formulae and compare them for both:

How do we get formulae for hyperbola if we know the formulae for ellipse?

• Tangent at (x₁, y₁): 

In case of hyperbola, the equation is xx₁/a² – yy₁/b² = 1 i.e. T = 0. The equation of tangent in case of ellipse is xx₁/a² + yy₁/b² = 1. Hence, again the only difference is of negative b².

• Equation of tangent in slope form:

?In terms of ‘m’, the equation of tangent to a hyperbola is y = mx + √(a²m² – b²), while in case of ellipse, the equation is y = mx + √(a²m² + b²).

• Equation of the normal at (x₁, y₁): 

In case of ellipse it is a²x/x₁ – b²y/y₁ = a²-b², and for hyperbola, the equation becomes a²x/x₁ + b²y/y₁ = a²-b².

• Equation of pair of tangents drawn from point (x₁, y₁): 

Pair of tangents to the hyperbola x²/a² – y²/b² = 1 is given by SS₁ = T², where S = x²/a² – y²/b² = 1, S₁ = x₁²/a² – y₁²/b² = 1 and T = xx₁/a² – yy₁/b² = 1

• Chord of Contact:

The chord of contact of tangents from (x₁, y₁) to the hyperbola x²/a² – y²/b² = 1 is given by T = 0 i.e. xx₁/a² – yy₁/b² = 1 and by changing b² to -b², we get the equation of chord of contact of tanegnts to an ellipse as xx₁/a² + yy₁/b² = 1.

• Polar: 

The Polar of the Pole (x₁, y₁) to the hyperbola x²/a² – y²/b² = 1 is given by T = 0 i.e. xx₁/a² – yy₁/b² = 1.

• Chord whose middle point is given: 

?The equation of Chord of hyperbola x²/a² – y²/b² = 1 whose middle point is (x₁, y₁) is given by T = S₁ i.e. x₁²/a² – y₁²/b² = xx₁/a² – yy₁/b² = 1.

You may refer the following video for more on hyperbola and ellipse

Question:

Can we represent a hyperbola mathematically in form of one parameter and is there any geometrical significance of that parameter like eccentric angle in case of ellipse?

Answer:

The simple answer to this question is yes. Before that, let us understand the concept of the auxiliary circle of a hyperbola. The circle described on the transverse axis of hyperbola as its diameter is called its auxiliary circle.

Hence, the equation of the auxiliary circle, described on AA’ as diameter, is

(x – a) [x – (–a)] + (y – 0)] + (y – 0)(y – 0) = 0

or x² + y² = a²

Now let us draw the foot N of any ordinate NP of the hyperbola draw a tangent NU to this circle, and join CU. Then 

CU = CN cos NCU

i.e.    x = CN = a sec NCU

The angle NCU is therefore the angle ∅.

Also NU = CU tan ∅ = a tan ∅

So that NP : NU = b : a

So the ordinate of the hyperbola is therefore in a constant ratio to the length of the tangent drawn from its foot to the auxiliary circle.

When it is desirable to express the co-ordinates of any point of the curve in terms of one parameter then we use x = a sec ∅, and y = b tan ∅. 

This angle ∅ is not so important an angle for the hyperbolas the eccentric angle is for the ellipse.

Question:

A tangent to the hyperbola x²/a² – y²/b² = 1 cuts the ellipse x²/a² + y²/b² = 1 in points L and R. Find the locus of the mid-point of LR.

Solution:

Let A(x₁, y₁) be the mid-point of the chord LR of the ellipse x²/a² + y²/b² = 1.

The equation of LR is xx₁/a² + yy₁/b² = x₁²/a² + y₁²/b² 

This gives y = -b²xx₁/ a²y₁ + b²/y₁(x₁²/a² + y₁²/b²)

This is a tangent to the hyperbola x²/a² – y²/b² = 1 if 

b⁴/y₁² (x₁²/a² + y₁²/b²)² = a²b⁴x₁²/ a⁴y₁² – b²

Hence, we have (x₁²/a² + y₁²/b²)² = x₁²/a² – y₁²/b²

Hence, the locus of (x₁, y₁) is (x²/a² + y²/b²)² = x²/a² – y²/b².
 

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