Commerce Exam  >  Commerce Notes  >  Mathematics (Maths) Class 11  >  Hyperbola

Hyperbola | Mathematics (Maths) Class 11 - Commerce PDF Download

HYPERBOLA

The Hyperbola is a conic whose eccentricity is greater than unity. (e > 1).

A. Standard Equation & Definition(s) 

Standard equation of the hyperbola is

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1, where b2 = a2(e2 - 1)

or a2 e2 = a2 + b2 i.e. e2 = 1 + Hyperbola | Mathematics (Maths) Class 11 - Commerce

= 1 + Hyperbola | Mathematics (Maths) Class 11 - Commerce 

 

(a) Foci : S Hyperbola | Mathematics (Maths) Class 11 - Commerce (ae, 0) & S' Hyperbola | Mathematics (Maths) Class 11 - Commerce (- ae, 0).

(b) Equations of directrices :Hyperbola | Mathematics (Maths) Class 11 - Commerce

(c) Vertices : A Hyperbola | Mathematics (Maths) Class 11 - Commerce (a, 0) & A' Hyperbola | Mathematics (Maths) Class 11 - Commerce (- a, 0).

(d) Latus rectum :

(i) Equation : x = ± ae

(ii) Length = Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce = 2a (e2 - 1) = 2e (distance from focus to directrix)

(iii) Ends : Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

(e) (i) Transverse Axis : The line segment A'A of length 2a in which the foci S' & S both lie is called the Transverse Axis of the Hyperbola.

(ii) Conjugate Axis : The line segment B'B between the two points B' ≡ (0, -b) & B ≡ (0, b) is called as the Conjugate Axis of the Hyperbola.

The Transverse Axis & the Conjugate Axis of the hyperbola are together called the Principal axes of the hyperbola.

(f) Focal Property : The difference of the focal distances of any point on the hyperbola is constant and equal to transverse axis i.e. Hyperbola | Mathematics (Maths) Class 11 - Commerce = 2a. The distance SS' = focal length.

(g) Focal distance : Distance of any point P(x, y) on Hyperbola from foci PS = ex - a & PS' = ex + a.

Ex.1 Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity Hyperbola | Mathematics (Maths) Class 11 - Commerce.

Sol. Let P(x, y) be any point on the hyperbola and PM is perpendicular from P on the directrix.

Then by definition SP = e PM    ⇒     (SP)2 = e2 (PM)2     ⇒      (x - 1)2 + (y - 2)2 = 3Hyperbola | Mathematics (Maths) Class 11 - Commerce

⇒ 5(x2 + y2 - 2x - 4y + 5) = 3(4x2 + y2 + 1 + 4xy - 2y - 4x)

⇒ 7x2 - 2y2 + 12 xy - 2x + 14y - 22 = 0 which is the required hyperbola.

Ex.2 The eccentricity of the hyperbola 4x2 - 9y2 - 8x = 32 is

Sol.

4x2 - 9y2 - 8x = 32   

⇒  4(x - 1)2 - 9y2 = 36      

⇒  Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1

Here a2 = 9, b2 = 4

Hyperbola | Mathematics (Maths) Class 11 - Commerce eccentricity e = Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

Ex.3 If foci of a hyperbola are foci of the ellipse Hyperbola | Mathematics (Maths) Class 11 - Commerce + Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1. If the eccentricity of the hyperbola be 2, then its equation is

Sol. For ellipse e = , so foci = (± 4, 0) 

For hyperbola

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hence equation of the hyperbola is Hyperbola | Mathematics (Maths) Class 11 - Commerce- Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1

Ex.4 Find the coordinates of foci, the eccentricity and latus-rectum, equations of directrices for the hyperbola 9x2 - 16y2 - 72x + 96y - 144 = 0 

Sol.

Equation can be rewritten as Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 so a = 4, b = 3 

b2 = a2(e2 - 1) given e = Hyperbola | Mathematics (Maths) Class 11 - Commerce

 Foci : X = ± ae, Y = 0 gives the foci as (9, 3), (-1, 3)

Centre : X = 0, Y = 0 i.e. (4, 3)

Directrices : X = ± i.e. x - 4 = Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce directrices are 5x - 36 = 0; 5x - 4 = 0 

Latus-rectum  Hyperbola | Mathematics (Maths) Class 11 - Commerce

B. Conjugate Hyperbola

Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & the transverse axes of the other are called Conjugate Hyperbolas of each other.

eg. Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 & -Hyperbola | Mathematics (Maths) Class 11 - Commerce + Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 are conjugate hyperbolas of each other.


Note :

(i) If e1 & e2 are the eccentricities of the hyperbola & its conjugate then e1-2 + e2-2 = 1.

(ii) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square.

(iii) Two hyperbolas are said to be similar if they have the same eccentricity.

Ex.5 The eccentricity of the conjugate hyperbola to the hyperbola x2 - 3y2 = 1 is

Sol.

Equation of the conjugate hyperbola to the hyperbola x2 - 3y2 = 1   is     -x2 + 3y2 = 1

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Here a2 = 1, b2 = 1/3    

Hyperbola | Mathematics (Maths) Class 11 - Commerce  eccentricity e = Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

C. Rectangular or Equilateral Hyperbola

The particular kind of hyperbola in which the lengths of the transverse & conjugate axis are equal is called an Equilateral Hyperbola. Note that the eccentricity of the rectangular hyperbola is Hyperbola | Mathematics (Maths) Class 11 - Commerce and the length of it's latus rectum is equal to it's transverse or conjugate axis.

D. Auxiliary Circle

A circle drawn with centre C & T.A. as a diameter is called the Auxiliary Circle of the hyperbola. Equation of the auxiliary circle is x2 + y2 = a2.

 Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

Note from the figure that P & Q are called the "Corresponding Points" on the hyperbola & the auxiliary circle `θ' is called the eccentric angle of the point `P' on the hyperbola. Hyperbola | Mathematics (Maths) Class 11 - Commerce

Parametric Equation : The equations x = a secθ & y = b tanθ together represents the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 where θ is a parameter. The parametric equations ; x = a cos h φ, y = b sin h φ also represents the same hyperbola.

General Note : Since the fundamental equation to the hyperbola only differs from that to the ellipse in having -b2 instead of b2 it will be found that many propositions for the hyperbola are derived from those for the ellipse by simply changing the sign of b2.

E. Position of a point `p' w.r.t a Hyperbola 

The quantity Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 is positive, zero or negative according as the point (x1, y1) lies outside, upon or within the curve.

 F. Line and a Hyperbola

The straight line y = mx + c is a secant, a tangent or passes outside the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce- Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 according as : c2 > = < a2m2 - b2.

Equation of a chord of the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 joining its two points P(α) & Q(β) is

Hyperbola | Mathematics (Maths) Class 11 - Commerce cos Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce sin Hyperbola | Mathematics (Maths) Class 11 - Commerce = cos Hyperbola | Mathematics (Maths) Class 11 - Commerce


Ex.6 Show that the line x cos α + y sin α = p touches the hyperbola - = 1 if a2 cos2 α - b2 sin2 α = p2.

Sol. The given line is x cos α + y sin α = p   ⇒    y sin α = -x cos α + p     ⇒   y = - x cot α + p cosec α Comparing this line with y = mx + c where m = - cot α, c = p cosec α

Since the given line touches the hyperbola - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 then c2 = a2m2 - b2    

⇒  p2 cosec2 α = a2 cot2 α - b2 or p2 = a2 cos2 α - b2 sin2 α

 

Ex.7 If (a sec θ, b tanθ) and (a secφ, b tanφ) are the ends of a focal chord of Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1, then tanHyperbola | Mathematics (Maths) Class 11 - Commerce tan Hyperbola | Mathematics (Maths) Class 11 - Commerce equal to

Sol. Equation of chord connecting the points (a sec θ, b tanθ) and (a sec φ, b tan φ) is

Hyperbola | Mathematics (Maths) Class 11 - Commerce cos Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - CommercesinHyperbola | Mathematics (Maths) Class 11 - Commerce = cos Hyperbola | Mathematics (Maths) Class 11 - Commerce                           .........(i)

If it passes through (ae, 0); we have, e cosHyperbola | Mathematics (Maths) Class 11 - Commerce = cos Hyperbola | Mathematics (Maths) Class 11 - Commerce

⇒ e = Hyperbola | Mathematics (Maths) Class 11 - Commerce   

⇒ tanHyperbola | Mathematics (Maths) Class 11 - Commerce.tanHyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce

Similarly if (i) passes through (-ae, 0), tan. tan = Hyperbola | Mathematics (Maths) Class 11 - Commerce

G. Tangent to the Hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 

(a) Point form : Equation of the tangent to the given hyperbola at the point (x1, y1) is Hyperbola | Mathematics (Maths) Class 11 - Commerce.

Note : In general two tangents can be drawn from an external point (x1, y1) to the hyperbola and they are y - y1 = m1 (x - x1) & y - y1 = m2 (x - x2), where m1 & m2 are roots of the equation (x12 - a2) m2 - 2x1y1 m + y12 + b2 = 0. If D < 0, then no tangent can be drawn from (x1 y1) to the hyperbola.

(b) Slope form : The equation of tangents of slope m to the given hyperbola is y = mx ± Hyperbola | Mathematics (Maths) Class 11 - Commerce.

Point of contact are Hyperbola | Mathematics (Maths) Class 11 - Commerce

Note : There are two parallel tangents having the same slope m.

(c) Parametric form : Equation of the tangent to the given hyperbola at the point (a sec θ, b tanθ) is Hyperbola | Mathematics (Maths) Class 11 - Commerce.

Note : Point of intersection of the tangents at θ1 & θ2 is x = aHyperbola | Mathematics (Maths) Class 11 - Commerce, y = b tan Hyperbola | Mathematics (Maths) Class 11 - Commerce

Ex.8 Find the equation of the tangent to the hyperbola x2 - 4y2 = 36 which is perpendicular to the line x - y + 4 = 0

Sol. Let m be the slope of the tangent. Since the tangent is perpendicular to the line x - y = 0

Hyperbola | Mathematics (Maths) Class 11 - Commerce m × 1 = - 1 ⇒ m = - 1

Since x2 - 4y2 = 36 or Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1. Comparing this with Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce a2 = 36 and b2 = 9

So the equation of tangents are y = (-1)x ± Hyperbola | Mathematics (Maths) Class 11 - Commerce

⇒ y= - x ± ⇒ x + y ± Hyperbola | Mathematics (Maths) Class 11 - Commerce = 0

Ex.9 The locus of the point of intersection of two tangents of the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce if the product of their slopes is c2, will be

Sol. Equation of any tangent of the hyperbola with slope m is y = mx ± Hyperbola | Mathematics (Maths) Class 11 - Commerce

If it passes through (x1, y1) then (y1 - mx1)2 = a2m2 - b2   

⇒      (x12 - a2) m2 - 2x1y1m + (y12 + b2) = 0

If m = m1, m2 then as given m1m2 = c2      ⇒    Hyperbola | Mathematics (Maths) Class 11 - Commerce= c2.

Hence required locus will be y2+b2 = c2(x- a2)

Ex.10 A common tangent to 9x2 - 16y2 = 144 and x2 + y2 = 9 is

Sol.

Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1, x2 + y2 = 9

Equation of tangent y = mx + Hyperbola | Mathematics (Maths) Class 11 - Commerce (for hyperbola)

Equation of tangent y = m'x + 3Hyperbola | Mathematics (Maths) Class 11 - Commerce (circle)

For common tangent m = m' and 3Hyperbola | Mathematics (Maths) Class 11 - Commerceor 9 + 9m= 16m2 - 9    or    7m2 = 18 ⇒ m=±3Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce required equation is Hyperbola | Mathematics (Maths) Class 11 - Commerce      or        Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

H. Normal to the Hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 

(a) Point form : The equation of the normal to the given hyperbola at the point P (x1, y1) on it is Hyperbola | Mathematics (Maths) Class 11 - Commerce + Hyperbola | Mathematics (Maths) Class 11 - Commerce = a2 - b2 = a2 e2.

(b) Slope form : The equation of normal of slope m to the given hyperbola is y = mx Hyperbola | Mathematics (Maths) Class 11 - Commerce foot of normal are Hyperbola | Mathematics (Maths) Class 11 - Commerce

(c) Parametric form : The equation of the normal at the point P (a sec θ, b tanθ) to the given hyperbola is Hyperbola | Mathematics (Maths) Class 11 - Commerce + Hyperbola | Mathematics (Maths) Class 11 - Commerce = a2 + b2 = a2 e2.

Ex.11 Line x cosα + y sinα = p is a normal to the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1, if

Sol. Equation of a normal to the hyperbola is ax cosθ + by cot θ = a2 + b2 comparing it with the given line equation

Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce    

⇒ sec θ = Hyperbola | Mathematics (Maths) Class 11 - Commerce,     

tanθ = Hyperbola | Mathematics (Maths) Class 11 - Commerce

Eliminating θ  , we get Hyperbola | Mathematics (Maths) Class 11 - Commerce-Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1

⇒ a2 sec2 α - b2 cosec2 α = Hyperbola | Mathematics (Maths) Class 11 - Commerce

 

Ex.12 The normal to the hyperbola = 1 meets the axes in M and N, and lines MP and NP are drawn at right angles to the axes. Prove that the locus of P is hyperbola (a2x2 - b2y2) = (a2 + b2)2.

Sol. Equation of normal at any point Q is ax cosθ + by cot θ = a2 + b2

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce M Hyperbola | Mathematics (Maths) Class 11 - CommerceHyperbola | Mathematics (Maths) Class 11 - Commerce, N Hyperbola | Mathematics (Maths) Class 11 - CommerceHyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce Let P Hyperbola | Mathematics (Maths) Class 11 - Commerce (h, k)

⇒     Hyperbola | Mathematics (Maths) Class 11 - Commerce  Hyperbola | Mathematics (Maths) Class 11 - Commerce

⇒        Hyperbola | Mathematics (Maths) Class 11 - Commerce         

Hyperbola | Mathematics (Maths) Class 11 - Commerce     locus of P is (a2x2 - b2y2) = (a2 + b2).

I. Highlights on Tangent and Normal

(a) Locus of the foot of the perpendicular drawn from focus of the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 upon any tangent is its auxilliary circle i.e. x2 + y2 = a2 & the product of lengths these perpendiculars is b2 (semi Conjugate Axis)2

(b) The portion of the tangent between the point of contact & the directrix subtends a right angle at the corresponding focus.

(c) The tangent & normal at any point of a hyperbola bisect the angle between the focal radii. This spells the reflection property of the hyperbola as "An incoming light ray" aimed towards one focus is reflected from the outer surface of the hyperbola towards the other focus. It follows that if an ellipse and a hyperbola have the same foci, they cut at right angles at any of their common point.

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Note that : the ellipse Hyperbola | Mathematics (Maths) Class 11 - Commerce= 1 & the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 (a > k > b > 0) are confocal and therefore orthogonal.

(d) The foci of the hyperbola and the points P and Q in which any tangent meets the tangents at the vertices are concyclic with PQ as diameter of the circle.

J. Director Circle

The locus of the intersection of tangents which are at right angles is known as the Director Circle of the hyperbola. The equation of the director circle is : x2 + y2 = a2 - b2.

If b2 < a2 this circle is real ; if b2 = a2 the radius of the circle is zero & it reduces to a point circle at the origin. In this case the centre is the only point from which the tangents at right angles can be drawn to the curve.

If b2 > a2, the radius of the circle is imaginary, so that there is no such circle & so no tangents at right angle can be drawn to the curve.

Note : Equation of chord of contact, chord with a given middle point, pair to tangents from an external point are to be interpreted in the similar way as in ellipse.

K. Asymptotes

Definition : If the length of the perpendicular let fall from a point on a hyperbola to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola, then the straight line is called the Asymptote of the Hyperbola.

To find the asymptote of the hyperbola :

Let y = mx + c is the asymptote of the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1.

Solving these two we get the quadratic as (b2 - a2m2) x2 - 2a2 mcx - a2(b2 + c2) = 0          ........(1)

In order that y = mx + c be an asymptote, both roots of equation (1) must approach infinity, the conditions for which are : coeff of x2 = 0 & coeff of x = 0.

Hyperbola | Mathematics (Maths) Class 11 - Commerce  

⇒ b2 - a2m2 = 0   or 

 Hyperbola | Mathematics (Maths) Class 11 - Commerce

a2 mc = 0 

⇒ c = 0.

Hyperbola | Mathematics (Maths) Class 11 - Commerce equations of asymptote are Hyperbola | Mathematics (Maths) Class 11 - Commerce and Hyperbola | Mathematics (Maths) Class 11 - Commerce combined equation to the asymptotes Hyperbola | Mathematics (Maths) Class 11 - Commerce 

Particular Case :

When b = a the asymptotes of the rectangular hyperbola. x2 - y2 = a2 are y = ± x which are at right angles.

Note :

(i) Equilateral hyperbola ⇔ rectangular hyperbola.

(ii) If a hyperbola is equilateral then the conjugate hyperbola is also equilateral

(iii) A hyperbola and its conjugate have the same asymptote.

(iv) The equation of the pair of asymptotes differ the hyperbola & the conjugate hyperbola by the same constant only.

(v) The asymptotes pass through the centre of the hyperbola & the bisectors of the angles between the asymptotes are the axes of the hyperbola.

(vi) The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.

(vii) Asymptotes are the tangent to the hyperbola from the centre.

(viii) A simple method to find the co-ordinates of the centre of the hyperbola expressed as a general equation of degree 2 should be remembered as : Let f(x, y) = 0 represents a hyperbola.

Find Hyperbola | Mathematics (Maths) Class 11 - Commerce & Hyperbola | Mathematics (Maths) Class 11 - Commerce. Then the point of intersection of Hyperbola | Mathematics (Maths) Class 11 - Commerce = 0 & Hyperbola | Mathematics (Maths) Class 11 - Commerce = 0 gives the centre of the hyperbola.

Ex.13 Find the asymptotes of the hyperbola 2x2 + 5xy + 2y2 + 4x + 5y = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.

Sol.

Let 2x2 + 5xy + 2y2 + 4x + 5y + λ = 0 be asymptotes. This will represent two straight line so

Hyperbola | Mathematics (Maths) Class 11 - Commerce λ = 2

⇒ 2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0 are asymptotes

⇒ (2x + y + 2) = 0 and (x + 2y + 1) = 0 are asymptotes

and 2x2 + 5xy + 2y2 + 4x + 5y + c = 0 is general equation of hyperbola.

Ex.14 Find the hyperbola whose asymptotes are 2x - y = 3 and 3x + y - 7 = 0 and which passes through the point (1, 1).

Sol. The equation of the hyperbola differs from the equation of the asymptotes by a constant

⇒ The equation of the hyperbola with asymptotes 3x + y - 7 = 0 and 2x - y = 3 is

(3x + y - 7)(2x - y - 3) + k = 0. It passes through (1, 1) ⇒k = -6.

Hence the equation of the hyperbola is (2x - y - 3) (3x + y - 7) = 6.

L. Highlights on Asymptotes

(a) If from any point on the asymptote a straight line be drawn perpendicular to the transverse axis, the product of the segments of this line, intercepted between the point & the curve is always equal to the square of the semi conjugate axis.

(b) Perpendicular from the foci on either asymptote meet it in the same points as the corresponding directrix & the common points of intersection lie on the auxiliary circle.

(c) The tangent at any point P on a hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 with centre C, meets the asymptotes in Q and R and cuts off a ΔCQR of constant area equal to ab from the asymptotes & the portion of the tangent intercepted between the asymptote is bisected at the point of contact. This implies that the locus of centre of the circle circumscribing the ΔCQR in case of a rectangular hyperbola is the hyperbola itself & for a standard hyperbola the locus would be the curve,

4(a2x2 - b2y2) = (a2 + b2)2.

(d) If the angle between the asymptote of a hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1 is 2θ then the eccentricity of the hyperbola is secθ.

M. Rectangular Hyperbola

Rectangular hyperbola referred to its asymptotes as axis of coordinates.

Hyperbola | Mathematics (Maths) Class 11 - Commerce

(a) Equation is xy = c2 with parametric representation

Hyperbola | Mathematics (Maths) Class 11 - Commerce

(b) Equation of a chord joining the points P(t1) & Q(t2) is x + t1 t2 y = c (t1 + t2) with slope m = Hyperbola | Mathematics (Maths) Class 11 - Commerce

(c) Equation of the tangent at P (x1, y1) is Hyperbola | Mathematics (Maths) Class 11 - Commerce + Hyperbola | Mathematics (Maths) Class 11 - Commerce = 2 & at P(t) is Hyperbola | Mathematics (Maths) Class 11 - Commerce + ty = 2c.

(d) Equation of normal is y - = t2 (x - ct)

(e) Chord with a given middle point as (h, k) is kx + hy = 2hk.

Note : For the hyperbola xy = c2

(i) Vertices : (c, c) & (-c, -c).

(ii) Foci : Hyperbola | Mathematics (Maths) Class 11 - Commerce

(iii) Directrices : x + y = ± Hyperbola | Mathematics (Maths) Class 11 - Commerce

(iv) Latus rectum : l = Hyperbola | Mathematics (Maths) Class 11 - Commerce = T . A = C . A

Ex.15 A triangle has its vertices on a rectangular hyperbola. Prove that the orthocentre of the triangle also lies on the same hyperbola.

Sol. Let t1, t2 and t3 are the vertices of the triangle ABC, described on the rectangular hyperbola xy = c2.

Hyperbola | Mathematics (Maths) Class 11 - Commerce co-ordinates of A, B and C are Hyperbola | Mathematics (Maths) Class 11 - Commerce respectively

Hyperbola | Mathematics (Maths) Class 11 - Commerce

Now slope of BC is Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce Slope of AD is t2t3

Equation of altitude AD is y - Hyperbola | Mathematics (Maths) Class 11 - Commerce = t2t3(x - ct1)

or t1y - c = xt1t2t3 - Hyperbola | Mathematics (Maths) Class 11 - Commerce .........(i)

Similarly equation of altitude BE is

t2y - c = xt1t2t3 - Hyperbola | Mathematics (Maths) Class 11 - Commerce .........(ii)

Solving (i) and (ii), we get the orthocentre Hyperbola | Mathematics (Maths) Class 11 - Commerce which lies on xy = c2.

Ex.16 Chords of the circle x2 + y2 = a2 touches the hyperbola x2/a2 - y2/b2 = 1. Prove that locus of their middle point is the curve (x2 + y2)2 = a2x2 - b2y2.

Sol. Let (h, k) be the mid-point of the chord of the circle x2 + y2 = a2,

so that its equation by T = S1 is     hx + ky = h2 + k2      or      Hyperbola | Mathematics (Maths) Class 11 - Commerce 

i.e. of the form y = mx + c. It will touch the hyperbola if c2 = a2m2 - b2

Hyperbola | Mathematics (Maths) Class 11 - CommerceHyperbola | Mathematics (Maths) Class 11 - Commerce = a2Hyperbola | Mathematics (Maths) Class 11 - Commerce or (h2 + k2)2 = a2h2 - b2k2

Generalizing, the locus of mid-point (h, k) is (x2 + y2)2 = a2x2 - b2y2

Ex.17 C is the centre of the hyperbola Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1. The tangent at any point P on this hyperbola meets the straight lines bx - ay = 0 and bx + ay = 0 in the points Q and R respectively. Show that CQ . CR = a2 + b2.

Sol. P is (a cosθ, b tanθ)

Tangent at P is Hyperbola | Mathematics (Maths) Class 11 - Commerce - Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1.

It meets bx - ay = 0 i.e. Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce in Q       

Hyperbola | Mathematics (Maths) Class 11 - Commerce Q is Hyperbola | Mathematics (Maths) Class 11 - Commerce.

It meets bx + ay = 0 i.e. Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce in R.      

Hyperbola | Mathematics (Maths) Class 11 - Commerce R is Hyperbola | Mathematics (Maths) Class 11 - Commerce

Hyperbola | Mathematics (Maths) Class 11 - Commerce CQ.CR = Hyperbola | Mathematics (Maths) Class 11 - Commerce    

Hyperbola | Mathematics (Maths) Class 11 - Commerce sec2q - tan2q = 1

Ex.18 A circle of variable radius cuts the rectangular hyperbola x2 - y2 = 9a2 in points P, Q, R and S. Determine the equation of the locus of the centroid of triangle PQR.

Sol. Let the circle be (x - h)2 + (y - k)2 = r2 where r is variable. Its intersection with x2 - y2 = 9a2 is obtained by putting y2 = x2 - 9a2.

x2 + x2 - 9a2 - 2hx + h2 + k2 - r2 = 2kHyperbola | Mathematics (Maths) Class 11 - Commerce

or [2x2 - 2hx + (h2 + k2 - r2 9a2)]2 = 4k2(x2 - 9a2) or 4x4 - 8hx3 + ....... = 0

Hyperbola | Mathematics (Maths) Class 11 - Commerce Above gives the abscissas of the four points of intersection. 

Hyperbola | Mathematics (Maths) Class 11 - Commerce    

⇒   x1 + x2 + x3 + x4 = 2h.

Similarly y1 + y2 + y3 + y4 = 2k

Now if (α,β) be the centroid of ΔPQR, then 3α = x1 + x2 + x3,  3β = y1 + y2 + y3

Hyperbola | Mathematics (Maths) Class 11 - Commerce x4 = 2h - 3a, y4 = 2k - 3β 

But (x4, y4) lies on x2 - y2 = 9a2  

Hyperbola | Mathematics (Maths) Class 11 - Commerce (2h - 3a)2 + (2k - 3b)2 = 9a2

Hence the locus of centroid (α,β) is (2h - 3x)2 + (2k - 3y)2 = 9a2 or Hyperbola | Mathematics (Maths) Class 11 - Commerce

Ex.19 If a circle cuts a rectangular hyperbola xy = c2 in A, B, C, D and the parameters of these four points be t1, t2 , t3 and t4 respectively, then prove that

(a) t1t2t3t4 = 1

(b) The centre of mean position of the four points bisects the distance between the centres of the two curves.

Sol. (a) Let the equation of the hyperbola referred to rectangular asymptotes as axes by xy = c2 or its parametric equation be

x = ct, y = c/t      ...........(i)

and that of the circle be x2 + y2 + 2gx + 2fy + k = 0   ...........(ii)

Solving (i) and (ii), we get

c2t2 + Hyperbola | Mathematics (Maths) Class 11 - Commerce + 2gct + 2fHyperbola | Mathematics (Maths) Class 11 - Commerce + k = 0    or  

c2t4 + 2gct3 + kt2 + 2f ct + c2 = 0       ...........(iii)

Above equation being of fourth degree in t gives us the four parameters t1, t2, t3, t4 of the points of intersection. 

Hyperbola | Mathematics (Maths) Class 11 - Commerce t1 + t2 + t3 + t4 = -Hyperbola | Mathematics (Maths) Class 11 - Commerce       ...........(iv)

t1t2t3 + t1t2t4 + t3t4t1 + t3t4t2 = Hyperbola | Mathematics (Maths) Class 11 - Commerce  ...........(v)

t1t2t3t4 = Hyperbola | Mathematics (Maths) Class 11 - Commerce = 1. It proves (a)   ...........(vi)

Dividing (v) by (vi), we get Hyperbola | Mathematics (Maths) Class 11 - Commerce    ...........(vii)

 

(b) The centre of mean position of the four points of intersection is

Hyperbola | Mathematics (Maths) Class 11 - Commerce = Hyperbola | Mathematics (Maths) Class 11 - Commerce, by (iv) and (vii) = (- g/2, - f/2)

Above is clearly the mid-point of (0, 0) and (- g, - f) i.e. the join of the centres of the two curves.

The document Hyperbola | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
All you need of Commerce at this link: Commerce
85 videos|243 docs|99 tests

Up next

Test: Hyperbola- 1
Test | 20 ques
Test: Hyperbola- 2
Test | 30 ques

FAQs on Hyperbola - Mathematics (Maths) Class 11 - Commerce

1. What is a hyperbola?
Ans. A hyperbola is a type of conic section that is formed by the intersection of a plane with two cones of opposite orientations. It consists of two separate curved branches that are symmetrical to each other.
2. How is a hyperbola different from other conic sections?
Ans. A hyperbola is different from other conic sections, such as ellipses and parabolas, because its eccentricity is greater than 1. This means that its branches are more spread out and do not form a closed curve like an ellipse.
3. What are the key properties of a hyperbola?
Ans. The key properties of a hyperbola include its foci, vertices, asymptotes, and eccentricity. The foci are two fixed points inside the hyperbola that determine its shape, while the vertices are the points where the hyperbola crosses the transverse axis. The asymptotes are the lines that the hyperbola approaches as it extends infinitely, and the eccentricity measures how stretched out the hyperbola is.
4. How can the equation of a hyperbola be determined?
Ans. The equation of a hyperbola can be determined using its center, foci, and other key properties. The general equation of a hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1 (for horizontal hyperbolas) or (y-k)^2/a^2 - (x-h)^2/b^2 = 1 (for vertical hyperbolas), where (h,k) represents the center of the hyperbola and a and b determine the shape and size.
5. In what real-life situations are hyperbolas used?
Ans. Hyperbolas have various real-life applications, such as in satellite communication, where the shape of the hyperbola helps determine the location of the satellite based on signal strengths. They are also used in optics to describe the shape of certain lenses and mirrors. Additionally, hyperbolas are utilized in navigation, radar systems, and orbital mechanics.
85 videos|243 docs|99 tests
Download as PDF

Up next

Test: Hyperbola- 1
Test | 20 ques
Test: Hyperbola- 2
Test | 30 ques
Explore Courses for Commerce exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Hyperbola | Mathematics (Maths) Class 11 - Commerce

,

Summary

,

shortcuts and tricks

,

past year papers

,

Semester Notes

,

Previous Year Questions with Solutions

,

Objective type Questions

,

video lectures

,

Important questions

,

mock tests for examination

,

pdf

,

study material

,

Sample Paper

,

practice quizzes

,

Hyperbola | Mathematics (Maths) Class 11 - Commerce

,

Exam

,

Hyperbola | Mathematics (Maths) Class 11 - Commerce

,

ppt

,

Viva Questions

,

MCQs

,

Extra Questions

,

Free

;