Hyperbola IIT JEE

I want the notes of Hyperbola Unit of Maths for preparation of IIT Joint Entrance Exam so can you provide me?

Hyperbola is an extremely important topic of IIT JEE Mathematics syllabus.

Students are advised to remember all the important properties of hyperbola on their fingertips.

Hyperbola notes-

Now, we know that hyperbola is a conic whose eccentricity is greater than unity i.e. e > 1.

The standard equation of hyperbola is x2/a2 – y2/b2 = 1, where b2 = a2 (e2-1).

This gives a2e2 = a2 + b2 or e2 = 1 + b2/a2 = 1 + (C.A / T.A.)2.
Various important terms and parameters of a hyperbola are listed below:

There are two foci of a hyperbola namely S (ae, 0) and S’(-ae, 0).

The equations of directrices are x = a/e and x = -a/e.

The latus rectum = 2b2/a = 2a (e2-1) and its length is therefore 2e.

The transverse and the conjugate axis of the hyperbola are together known as principal axis of the hyperbola.

For a hyperbola with the given equation x2/a2 – y2/b2 = 1, its conjugate is given by the equation – x2/a2 + y2/b2 = 1.

In parametric coordinates, the equations x = a sec θ and y = b tan θ together represent the hyperbola x2/a2 – y2/b2 = 1, where θ is a parameter. Another set of equations representing the same hyperbola are x = a cosh ∅ and y = b sinh ∅.

Some Important Results and Properties:

Focal Property: The difference of the focal distances of any point on the hyperbola is constant and is equal to transverse axis i.e. ||PS – PS’|| = 2a. The distance SS’ is the focal length.

If e1 and e2 are the eccentricities of the hyperbola and the conjugate hyperbola then this relation holds good: e1-2 + e2-2 = 1.

If the lengths of the transverse axis and the conjugate axis are equal then the hyperbola is said to be a rectangular or equilateral hyperbola.

The eccentricity of the rectangular hyperbola is √2 and the length of its latus rectum is same as its transverse or conjugate axis.

The point (x1, y1) lies within, on or outside the hyperbola if the quantity x12/a2 – y12/b2 = 1 is positive, zero or negative.

The line y = mx + c is a secant, a tangent or passes from the outside of the hyperbola according as c2 > = < a2m2 - b2.

Equations of tangents:

1. The equation of tangent to the hyperbola at the point (x1, y1) is xx1/a2 – yy1/b2 = 1.

2. In general, two tangents can be drawn to the hyperbola from an external point (x1, y1) to the hyperbola and they are given by the equations (y-y1) = m1(x-x1) and (y-y2) = m2(x-x2), where m1 and m2 are the roots of the equation (x12 – a2)m2 – 2x1y1m + y12 + b2 = 0.

3. In parametric coordinates, the equation of the tangent is given by x sec θ/a – y tan θ/b = 1.

4. Also, y = mx ± √(a2m2 – b2) can be taken as a tangent to the hyperbola x2/a2 – y2/b2 = 1.

5. The portion of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.

Equations of normals:

1. The equation of the normal to the hyperbola at the point (x1, y1) is a2x/x1 + b2y/y1 = a2-b2 = a2e2.

2. In parametric coordinates, the equation becomes ax /sec θ + by/ tan θ = a2 + b2 = a2e2.

The tangent and normal at any point of hyperbola bisect the angle between the focal radii.

The asymptotes of a hyperbola and its conjugate are the same.

Asymptotes are the tangents to the centre.

The asymptotes pass through the centre of the hyperbola and the axes of the hyperbola are the bisectors of the angles between the asymptotes.

Rectangular hyperbola:

1. The equation is xy = c2 with parametric representation x = ct, y = c/t, t ∈ R – {0}.

2. Equation of the tangent at P(x1, y1) is x/x1 + y/y1 = 2 and at P(t) is x/t + ty = 2.

3. Equation of normal is y – c/t = t2(x-ct).

The equation of pair of tangents drawn from a point P(x1, y1) to the hyperbola x2/a2 – y2/b2 = 1 is SS1 = T2, where S = x2/a2 – y2/b2 -1, S1 = x12/a2 – y12/b2 – 1, T = xx1/a2 – yy1/b2 - 1.

The straight line y = mx + c is a tangent to the curve, if c2 = a2m2 – b2. In other words, y = mx + √a2m2–b2 touches the curve for all those values of m when m > b/a or m < –b/a.
The equation of the chord through the points θ1 and θ2 is



The equation of the diameter bisecting the chords of slope m of the hyperbola x2/a2 – y2/b2 = 1 is y = b2/a2m.

The polar of the point P(x1, y1) with respect to the hyperbola x2/a2 – y2/b2 = 1 is T = 0, where T = xx1/a2 – yy1/b2 - 1.

Polar of the focus is the directrix.

Locus of the feet of the perpendicular drawn from focus of the hyperbola x2/a2 – y2/b2 = 1 upon any tangent is its auxiliary circle i.e. x2 + y2 = a2 and the product of the feet of these perpendiculars is b2.

A rectangular hyperbola circumscribing a triangle passes through the orthocentre of this triangle.

If a circle meets a rectangular hyperbola at four points, then the mean value of the points of intersection is the mid-point of the line joining the centres of circle and hyperbola.

Illustration:

For the hyperbola x2/cos2 θ – y2/ sin2 θ = 1, which of the following remains constant with change in a? (IIT JEE 2003)

(a) abcissae of vertices (b) abcissae of foci

(c) eccentricity (d) directrix

Solution:

The given equation is x2/cos2 θ – y2/ sin2 θ = 1.

Here, a2 = cos2 θ and b2 = sin2 θ.

Now, foci = (± ae, 0), where ae = √a2 + b2

Hence, foci are ( ±1, 0) whereas vertices are (± cos θ, 0).

eccentricity ae = 1 or e = 1/cos θ

Hence, this shows that focus remains constant with change in θ.

Solved Examples on Hyperbola

Show that the line 4x – 3y = 9 touches the hyperbola 4x2 – 9y2 = 27.

Solution:

We know that if the line y = mx + c touches the hyperbola x2/a2–y2/b2 = 1, then c2= a2m2 – b2
Here the hyperbola is

x2/(27/4)–y2/(27/9) = 1

i.e. here a2 = 27/4 b2 = 27/9 = 3

And comparing 4x – 3y = 9 with y = mx + c, we get

‘m’ = 4/3, ‘c’ = –3

∴ a2m2 – b2 = (27/4)(4/3)3 – 3 = 12 – 3 = 9 = (–3)2

or a2m2 – b2 = c2

Hence the given line touches the given hyperbola.

Example:

Prove that the mid points of chords of the hyperbola x2/a2–y2/b2 = 1 parallel to the diameter y = mx lie on the diameter a2my = b2x

Solution:

The hyperbola is x2/a2–y2/b2 = 1 …… (1)

The equation of any chord parallel to the diameter

y = mx is y = mx + c …… (2)

Eliminating y between (1) and (2), we get

x2/a2–(mx+c)2/b2 = 1

⇒ x2(b2 – a2m2) – 2a2m cx – a2(b2 + c2) = 0

⇒ b2x3 = a2my3

∴ The locus of (x3, y3) is

b2x = a2my Hence proved.

Example:

Prove that the angle subtended b any chord of a rectangular hyperbola at the centre is the supplement of the angle between the tangents at the end of the chord.

Solution:

Let the equation of the hyperbola be x2 – y2 = a2 and P and Q be any two points on it such that their coordinates are respectively (a sec ?1, a tan ?1) and (a sec ?2, a tan ?2) and C be the centre of the hyperbola.

Equation of the line PC is y – 0 = atan?1–0/asec?1–0 (x – 0)

⇒ y = x sin ?1 …… (1)

Similarly equation to QC will be y = x sin ?2 …… (2)

If a be the angle between PC and QC, then

tan α = sin?1–sin?2/1+sin?1sin?2 …… (3)

Again the equation to the tangent at P is

x a sec ?1– y a tan ?1 = a2

y = x/sin?1 – acos?1/sin?1 …… (4)

Similarly the equation to the tangent at Q2 is

y = x/sin?2 – a cos?2/sin?2

If b be the angle between the tangents at f1 and f2, then

tan ß = 1/sin?2 1/sin?2/1+1/sin?1 1/sin?2 = sin?2–sin?1/1+sin?1sin?2

= –(sin?1–sin?2)/1+(sin?1sin?2)

⇒ tanß = –tanα

⇒ tanß = tan (π – α) (By (3))

⇒ ß = π – α Hence proved.

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