Ellipse | Mathematics (Maths) for JEE Main & Advanced PDF Download

Standard Equation & Definition 

Standard equation of an ellipse referred to its principal axes along the co-ordinate axes is

Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1. where a > b & b2 = a2 (1 - e2) ⇒ a2 - b2 = a2 e2.

 Ellipse | Mathematics (Maths) for JEE Main & Advanced

where e = eccentricity (0 < e < 1).

FOCI : S Ellipse | Mathematics (Maths) for JEE Main & Advanced (ae, 0) & S' Ellipse | Mathematics (Maths) for JEE Main & Advanced (- ae, 0).

(a) Equation of directrices :

Ellipse | Mathematics (Maths) for JEE Main & Advanced

(b) Vertices :

A' Ellipse | Mathematics (Maths) for JEE Main & Advanced (-a, 0) & A Ellipse | Mathematics (Maths) for JEE Main & Advanced (a, 0)

(c) Major axis : The line segment A' A in which the foci S' S S lie is of length 2a & is called the major axis (a > b) of the ellipse. Point of intersection of major axis with directrix is called the foot of the  directrix (z)Ellipse | Mathematics (Maths) for JEE Main & AdvancedEllipse | Mathematics (Maths) for JEE Main & Advanced.

(d) Minor Axis : The y-axis intersects the ellipse in the points B' Ellipse | Mathematics (Maths) for JEE Main & Advanced (0, - b) & B Ellipse | Mathematics (Maths) for JEE Main & Advanced (0, b). The line segment B'B of length 2b (b < a) is called the Minor Axis of the ellipse.

(e) Principal Axes : The major & minor axis together are called Principal Axes of the ellipse.

(f) Centre : The point which bisects every chord of the conic drawn through it is called the centre of the conic. C Ellipse | Mathematics (Maths) for JEE Main & Advanced (0, 0) the origin is the centre of the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

(g) Diameter : A chord of the conic which passes through the centre is called a diameter of the conic.

(h) Focal Chord : A chord which passes through a focus is called a focal chord.

(i) Double Ordinate : A chord perpendicular to the major axis is called a double ordinate.

 (j) Latus Rectum : The focal chord perpendicular to the major axis is called the latus rectum.

(i) Length of latus rectum (LL') = Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced = 2a(1 - e2) = 2e

(ii) Equation of latus rectum : x = ± ae.

(iii) Ends of the latus rectum are Ellipse | Mathematics (Maths) for JEE Main & Advanced and L1'Ellipse | Mathematics (Maths) for JEE Main & Advanced.

 

(k) Focal radii : SP = a -ex and S'P = a + ex    ⇒     SP + S'P = 2a = Major axis.

(l) Eccentricity : e = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Note :

(i) The sum of the focal distances of any point on the ellipse is equal to the major axis. Hence distance of focus from the extremity of a minor axis is equal to semi major axis. i.e. BS = CA.

(ii) If the equation of the ellipse is given as Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 & nothing is mentioned, then the rule is to assume that a > b.

Ex.1 If LR of an ellipse is half of its minor axis, then its eccentricity is

Sol.

As given Ellipse | Mathematics (Maths) for JEE Main & Advanced = b   ⇒    2b = a 

⇒    4b2 = a⇒ 4a2(1 - e2) = a2   

⇒     1 - e2 = 1/4 Ellipse | Mathematics (Maths) for JEE Main & Advanced e = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ex.2 Find the equation of the ellipse whose foci are (2, 3), (-2, 3) and whose semi minor axis is of length Ellipse | Mathematics (Maths) for JEE Main & Advanced.

Sol. Here S is (2, 3) & S' is (-2, 3) and        Ellipse | Mathematics (Maths) for JEE Main & Advanced    

⇒ SS' = 4 = 2ae  ⇒ ae = 2

but b2 = a2 (1 - e2)  ⇒  5 = a2 - 4      

  ⇒    a = 3. Hence the equation to major axis is y = 3

Centre of ellipse is midpoint of SS' i.e. (0, 3)

Ellipse | Mathematics (Maths) for JEE Main & Advanced Equation to ellipse is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 or Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

Ex.3 Find the equation of the ellipse having centre at (1, 2), one focus at (6, 2) and passing through the point (4, 6).

Sol.

With centre at (1, 2), the equation of the ellipse is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1.

It passes through the point (4, 6)      ⇒          Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 .............(i)

Distance between the focus and the centre = (6 - 1) = 5 = ae

⇒ b2 = a2 - a2e2 = a2 - 25 .............(ii)

Solving for a2 and b2 from the equation (i) and (ii), we get a2 = 45 and b2 = 20.

Hence the equation of the ellipse is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

B. Another form of Ellipse

Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1, (a < b)

(a) AA' = Minor axis = 2a

(b) BB' = Major axis = 2b

(c) a2 = b2 (1 - e2)

(d) Latus rectum LL' = L1L1' = Ellipse | Mathematics (Maths) for JEE Main & Advanced.

equation y = ± be

(e) Ends of the latus rectum are :

Ellipse | Mathematics (Maths) for JEE Main & Advanced,

Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ellipse | Mathematics (Maths) for JEE Main & Advanced

(f) Equation of directrix y = ± Ellipse | Mathematics (Maths) for JEE Main & Advanced.

(g) Eccentricity : e = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ex.4 The equation of the ellipse with respect to coordinate axes whose minor axis is equal to the distance between its foci and whose LR = 10, will be

Sol. When a > b

As given 2b = 2ae  ⇒ b = ae       ............(i)

Also Ellipse | Mathematics (Maths) for JEE Main & Advanced = 10     ⇒b2 = 5a     . ...........(ii)

Now since b2 = a2 - a2e2  ⇒  b2 = a2 - b2  ⇒ 2b2 = a2 ............(iii) [From (i)]

(ii), (iii) ⇒ a2 = 100, b2 = 50

Hence equation of the ellipse will be Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 ⇒  x2 + 2y2 = 100

Similarly when a < b then required ellipse is 2x2 + y2 = 100

C. General Equation of an Ellipse

 Ellipse | Mathematics (Maths) for JEE Main & Advanced

Let (a, b) be the focus S, and lx + my + n = 0 is the equation of directrix.

Let P(x, y) be any point on the ellipse. Then by definition.

⇒ SP = e PM (e is the eccentricity)  

⇒    (x - a)2 + (y - b)2 = e2Ellipse | Mathematics (Maths) for JEE Main & Advanced

⇒  (l2 + m2) {(x - a)2 + (y - b)2} = e2{lx + my + n}2

D. Auxilliary Circle/eccentric Angle

 Ellipse | Mathematics (Maths) for JEE Main & Advanced

A circle described on major axis as diameter is called the auxilliary circle. Let Q be a point on the auxilliary circle x2 + y2 = a2 such that QP produced is perpendicular to the x-axis then P & Q are called as the corresponding points on the ellipse & the auxilliary circle respectively. `θ' is called the eccentric angle of the point P on the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced

Note that :Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Hence "If from each point of a circle perpendiculars are drawn upon a fixed diameter then the locus of the points dividing these perpendiculars in a given ratio is an ellipse of which the given circle is the auxilliary circle".

E. position of a point w.r.t an ellipse

The point P(x1, y1) lies outside, inside or on the ellipse according as ; Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1 > < or = 0.

F. Line and an ellipse

The line y = mx + c meets the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 in two points real, coincident or imaginary according as c2 is < = or > a2m2 + b2.

Hence y = mx + c is tangent to the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 if c2 = a2m2 + b2.

The equation to the chord of the ellipse joining two points with eccentric angles a & b is given by Ellipse | Mathematics (Maths) for JEE Main & Advanced cos Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced sin Ellipse | Mathematics (Maths) for JEE Main & Advanced = cos Ellipse | Mathematics (Maths) for JEE Main & Advanced.

G. Parametric representation 

The equations x = cosθ & y = b sinθ together represent the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

where θ is a parameter (eccentric angle).

Note that : If P(θ ) ≡ (a cosθ, b sinθ) is on the ellipse then ;

Q(θ ) ≡ (a cosθ, a sinθ) is on the auxilliary circle.

Ex.5 For what value of l does the line y = x + l touches the ellipse 9x2 + 16y2 = 144.

Sol.Ellipse | Mathematics (Maths) for JEE Main & Advanced Equation of ellipse is 9x2 + 16y2 = 144 or Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

Comparing this with Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 then we get a2 = 16 and b2 = 9

and comparing the line y = x + λ with y = mx + c          

Ellipse | Mathematics (Maths) for JEE Main & Advanced m = 1 and c = λ

If the line y = x + λ touches the ellipse 9x2 + 16y2 = 144, then c2 = a2m2 + b2

⇒ λ2 = 16 × 12 + 9          

⇒  λ2 = 25  

Ellipse | Mathematics (Maths) for JEE Main & Advanced λ = ± 5

Ex.6 If α, β are eccentric angles of end points of a focal chord of the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1, then tan α/2. tan β/2 is equal to

Sol.

Equation of line joining points `α' and `β' is cos Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced sin Ellipse | Mathematics (Maths) for JEE Main & Advanced = cos Ellipse | Mathematics (Maths) for JEE Main & Advanced

If it is a focal chord, then it passes through focus (ae, 0), so e cos Ellipse | Mathematics (Maths) for JEE Main & Advanced = cos Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced        

Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced        

⇒ tanEllipse | Mathematics (Maths) for JEE Main & Advanced tanEllipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced

 

H. Director Circle

Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle. The equation to this locus is x2 + y2 = a2 + b2 i.e. a circle whose centre is the centre of the ellipse & whose radius is the length of the line joining the ends of the major & minor axis.

Ex.7 A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x2 + 2y2 = 6 are at right angles.

Sol.

Given ellipse are Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1     ...........(i)

and, Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1    ...........(ii)

any tangent to (i) is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1   ...........(iii)

It cuts (ii) at P and Q, and suppose tangent at P and Q meet at (h, k) Then equation of chord of contact of (h, k) with respect to ellipse (ii) is   Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 ...........(iv)

comparing (iii) and (iv), we get Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1      

⇒ cosθ = Ellipse | Mathematics (Maths) for JEE Main & Advanced and sinθ = Ellipse | Mathematics (Maths) for JEE Main & Advanced 

 ⇒ h2 + k2 = 9

locus of the point (h, k) is x2 + y2 = 9

⇒ x2 + y2 = 6 + 3 = a2 + b2

i.e. director circle of second ellipse. Hence the tangents are at right angles.

I. Tangent to the Ellipse  Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 

(a) Point form : Equation of tangent to the given ellipse at its point (x1, y1) is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

Note :  For general ellipse replace x2 by xx1,y2by yy1,2x by x + x1,2y by y + y1,2xy by xy1+yx1 & c by c.

(b) Slope form : Equation of tangent to the given ellipse whose slope is `m', is y = mx ± Ellipse | Mathematics (Maths) for JEE Main & Advanced

Point of contact are Ellipse | Mathematics (Maths) for JEE Main & Advanced

Note : There are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.

(c) Parametric form : Equation of tangent to the given ellipse at its point (a cosθ, b sinθ), is

Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

Note :

(i) The eccentric angles of point of contact of two parallel tangents differ by π.

(ii) Point of intersection of the tangents at the point α & β is Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ex.8 Find the equations of the tangents to the ellipse 3x2 + 4y2 = 12 which are perpendicular to the line y + 2x = 4.

Sol. Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4.

Ellipse | Mathematics (Maths) for JEE Main & Advanced mx - 2 = - 1

m = 1/2

Since 3x2 + 4y2 = 12   or    Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1.

Comparing this with Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1            

Ellipse | Mathematics (Maths) for JEE Main & Advanceda2 = 4 and b2 = 3

So the equation of the tangent are Ellipse | Mathematics (Maths) for JEE Main & Advanced    

Ellipse | Mathematics (Maths) for JEE Main & Advanced 

or  

x - 2y ± 4 = 0. 

Ex.9 The tangent at a point P on an ellipse intersects the major axis in T and N is the foot of the perpendicular from P to the same axis. Show that the circle drawn on NT as diameter intersects the auxilliary circle orthogonally.

Sol.

Let the equation of the ellipse be Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1. Let (acosθ, bsinθ) be a point on the ellipse. The equation of the tangent at P is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1. It meets the major axis at T ≡ (a secθ, 0). The coordinates of N are (a cosθ, 0). The equation of the circle with NT as its diameter is (x - asecθ) (x - acosθ) + y2 = 0   ⇒  x2 + y2 - ax(secθ + cosθ) + a2 = 0

It cuts the axuilliary circle x2 + y2 - a2 = 0 orthogonally if 2g . 0 + 2f . 0 = a2 - a2 = 0. which is true.

J. Normal to the Ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

(a) Point form : Equation of the normal to the given ellipse at (x1, y1) is Ellipse | Mathematics (Maths) for JEE Main & Advanced - Ellipse | Mathematics (Maths) for JEE Main & Advanced = a2 - b2 = a2e2

(b) Slope form : Equation of a normal to the given ellipse whose slope is `m' is y = mx Ellipse | Mathematics (Maths) for JEE Main & Advanced

(c) Parametric form : Equation of the normal to the given ellipse at the point (acosθ, bsinθ) is ax. sec θ - by. cosec θ = (a2 - b2).

Ex.10 Find the condition that the line lx + my = n may be a normal to the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1.

Sol.

Equation of normal to the given ellipse at (a cosθ, b sinθ) is Ellipse | Mathematics (Maths) for JEE Main & Advanced - Ellipse | Mathematics (Maths) for JEE Main & Advanced = a2 - b2 ..........(i)

If the line lx + my = n is also normal to the ellipse then there must be a value of θ for which line (i) and line lx + my = n are identical. For the value of q we have

Ellipse | Mathematics (Maths) for JEE Main & Advanced or Ellipse | Mathematics (Maths) for JEE Main & Advanced cosθ = Ellipse | Mathematics (Maths) for JEE Main & Advanced ........(iii)

and sinθ = Ellipse | Mathematics (Maths) for JEE Main & Advanced .....(iv)

Squaring and adding (iii) and (iv), we get 1 = Ellipse | Mathematics (Maths) for JEE Main & AdvancedEllipse | Mathematics (Maths) for JEE Main & Advanced which is the required condition.

Ex.11 If the normal at an end of a latus-rectum of an ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 passes through one extremity of the minor axis, show that the eccentricity of the ellipse is given by e = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Sol. The co-ordinates of an end of the latus-rectum are (ae, b2/a). The equation of normal at P(ae, b2/a) is

Ellipse | Mathematics (Maths) for JEE Main & Advanced - Ellipse | Mathematics (Maths) for JEE Main & Advanced = a2 - b2 

or             

Ellipse | Mathematics (Maths) for JEE Main & Advanced - ay = a2 - b2

It passes through one extremity of the minor axis whose co-ordinates are (0, -b)

Ellipse | Mathematics (Maths) for JEE Main & Advanced 0 + ab = a2 - b2  

⇒ (a2b2) = (a2 - b2)2           

⇒    a2.a2(1 - e2) = (a2 e2)2

⇒ 1 - e2 + e4

⇒ e4 + e2 - 1 = 0    

⇒    (e2)2 + e2 - 1 = 0

Ellipse | Mathematics (Maths) for JEE Main & Advanced e2 = Ellipse | Mathematics (Maths) for JEE Main & Advanced ⇒ e = Ellipse | Mathematics (Maths) for JEE Main & Advanced (taking positive sign)

 

Ex.12 P and Q are corresponding points on the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 and the auxilliary circles respectively. The normal at P to the ellipse meets CQ in R, where C is the centre of the ellipse, Prove that CR = a + b

Sol. Let P Ellipse | Mathematics (Maths) for JEE Main & Advanced (acosθ, bsinθ)  

Ellipse | Mathematics (Maths) for JEE Main & AdvancedQ Ellipse | Mathematics (Maths) for JEE Main & Advanced (acosθ, asinθ)

Equation of normal at P is (asecq) x - (bcosecθ)y = a2 - b2    ............(i)

equation of CQ is y = tanθ . x  ............(ii)

Solving equation (i) & (ii), we get

(a - b)x = (a2 - b2) cosθ           

⇒     x = (a + b) cosθ, & y = (a + b) sinθ

Ellipse | Mathematics (Maths) for JEE Main & Advanced R Ellipse | Mathematics (Maths) for JEE Main & Advanced (a + b)cosθ, (a + b)sinθ  

Ellipse | Mathematics (Maths) for JEE Main & Advanced CR = a + b

K. Pair of tangents

Ellipse | Mathematics (Maths) for JEE Main & Advanced

If P(x1, y1) be any point lies outside the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1,

and a pair of tangents PA, PB can be drawn to it from P.

Then the equation of pair of tangents of PA and PB is SS1 = T2

where        S1 = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1,           T = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1

i.e. Ellipse | Mathematics (Maths) for JEE Main & AdvancedEllipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced

L. Chord of contact

If PA and PB be the tangents from point P(x1, y1) to the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1.

Then the equation of the chord of contact AB is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 or T = 0 (at x1, y1)

Ex.13 If tangents to the parabola y2 = 4ax intersect the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 at A and B, the find the locus of point of intersection of tangents at A and B.

Sol. Let P ≡ (h, k) be the point of intersection of tangents at A & B

Ellipse | Mathematics (Maths) for JEE Main & Advanced equation of chord of contact AB is Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1 ..............(i)

which touches the parabola.

Equation of tangent to parabola y2 = 4ax is y = mx + Ellipse | Mathematics (Maths) for JEE Main & Advanced         ⇒     mx - y = -Ellipse | Mathematics (Maths) for JEE Main & Advanced         ..............(ii)

equation (i) & (ii) as must be same

Ellipse | Mathematics (Maths) for JEE Main & AdvancedEllipse | Mathematics (Maths) for JEE Main & Advanced ⇒ m = -Ellipse | Mathematics (Maths) for JEE Main & AdvancedEllipse | Mathematics (Maths) for JEE Main & Advanced & m = Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ellipse | Mathematics (Maths) for JEE Main & Advanced -Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced ⇒ locus of P is y2 = - Ellipse | Mathematics (Maths) for JEE Main & Advanced.x

Ellipse | Mathematics (Maths) for JEE Main & Advanced

M. Equation of chord with mid point (x1, y1)

The equation of the chord of the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1, whose mid-point be (x1, y1) is T = S1

where T = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1, S1 = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1 i.e. Ellipse | Mathematics (Maths) for JEE Main & Advanced

 

Ex.14 Find the locus of the mid - point of focal chords of the ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1.

Sol. Let P Ellipse | Mathematics (Maths) for JEE Main & Advanced (h, k) be the mid - point

Ellipse | Mathematics (Maths) for JEE Main & Advanced

Ellipse | Mathematics (Maths) for JEE Main & Advanced equation of chord whose mid-point is given Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1 = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced - 1 since it is focal chord,

Ellipse | Mathematics (Maths) for JEE Main & Advanced It passes through focus, either (ae, 0) or (-ae, 0)

If it passes through (ae, 0)          

   Ellipse | Mathematics (Maths) for JEE Main & Advanced        locus is Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced

If it passes through (-ae, 0)      

      Ellipse | Mathematics (Maths) for JEE Main & Advanced          locus is - Ellipse | Mathematics (Maths) for JEE Main & Advanced = Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced

 

N. Important Highlights

Ellipse | Mathematics (Maths) for JEE Main & Advanced

Referring to an ellipse Ellipse | Mathematics (Maths) for JEE Main & Advanced + Ellipse | Mathematics (Maths) for JEE Main & Advanced = 1

(a) If P be any point on the ellipse with S & S'

as its foci then l(SP) + l(S'P) = 2a.

(b) The tangents & normal at a point P on the ellipse bisect the external & internal angles between the focal distances of P.

This refers to the well known reflection property of the ellipse which states that rays from one focus are reflected through other focus & vice versa. Hence we can deduce that the straight lines joining each focus to the foot of the perpendicular from the other focus upon the tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis.

(c) The product of the length's of the perpendicular segments from the foci on any tangent to the ellipse is b2 and the feet of these perpendiculars lie on its auxiliary circle and the tangents at these feet to the auxiliary circle meet on the ordinate of P and that the locus of their point of intersection is a similar ellipse as that of the original one.

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

(e) If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively, & if CF be perpendicular upon this normal, then

(i) PF . PG = b2

(ii) PF . Pg = a

(iii) PG . Pg = SP . S'P

(iv) CG . CT = CS2

(v) locus of the mid point of Gg is another ellipse having the same eccentricity as that of the original ellipse.

[where S and S' are the focii of the ellipse and T is the point where tangent at P meet the major axis]

(f) The circle on any focal distance as diameter touches the auxilliary circle.

(g) Perpendiculars from the centre upon all chords which join the ends of any perpendicular diameters of the ellipse are of constant length.

(h) If the tangent at the point P of a standard ellipse meets the axes in T and t and CY is the perpendicular on it form the centre then,

(i) T t . PY = a2 - b2  and          
(ii) least value of T t is a + b.

The document Ellipse | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on Ellipse - Mathematics (Maths) for JEE Main & Advanced

1. What is an ellipse in geometry?
Ans. An ellipse is a closed curve that is shaped like an elongated circle. It is defined as the set of all points in a plane, the sum of whose distances from two fixed points (foci) is constant.
2. How is the length of the major and minor axes of an ellipse calculated?
Ans. The length of the major axis is the distance between the two furthest points on the ellipse, passing through the center. The length of the minor axis is the distance between the two points on the ellipse that are perpendicular to the major axis and pass through the center.
3. What is the eccentricity of an ellipse?
Ans. The eccentricity of an ellipse is a measure of how elongated the ellipse is. It is calculated as the ratio of the distance between the center and one of the foci to the distance between the center and a point on the ellipse.
4. How is the area of an ellipse calculated?
Ans. The area of an ellipse is calculated using the formula A = π * a * b, where 'a' and 'b' are the lengths of the semi-major and semi-minor axes of the ellipse, respectively.
5. Can an ellipse have a negative area?
Ans. No, an ellipse cannot have a negative area. The area of an ellipse is always a positive value, as it represents the space enclosed by the curve.
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