General Equation of an Ellipse

The general equation of an ellipse with major axis as the x- axis is

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2)= 1`

where (h,k) is the centre of the ellipse.

ellipse with major axis as x - axis

ellipse

The general equation of an ellipse with major axis as the y-axis is

`((x-h)^2)/(b^2) + ((y-k)^2)/(a^2)= 1`

ellipse with major axis as y - axis

ellipse

Lets solve some problems to find the general equation of ellipse.

Find the General Equation of an Ellipse - Examples

Ex : 1 Find the general equation of an ellipse whose vertices are (-1,4) and (-7,4) and eccentricity is `1/3` .

Sol : From the given data the major axis is parallel to x- axis.

`:.` The general equation form of an equation is `((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) =1`

The centre is the midpoint of AA'

example 1 to find general equation

`:.` Cis ( `(-1-7)/2,(4+4)/2` ) = (-4,4)

Thus the equation becomes

`((x+4)^2)/(a^2) +( (y-4)^2)/(b^2) = 1`

We know that AA' = 2a = 6 `rArr` a=3

b2 = a2(1-e2) = 9( 1- `1/9` ) = 8

The required equation is `((x+4)^2)/9 + ((y-4)^2)/8 = 1`

Ex 2: Find the equation of the ellipse whose foci are (1,3) and (1,9) and eccentricity is `1/2`

Sol : From the given data the major axis is parallel to y-axis

`:.` the general equation is of the form

`((x-h)^2)/(b^2) + ((y-k)^2)/(a^2) = 1`

example 2 to find general equation

The centre of the ellipse is the midpoint of F1F2

`:.` C is ( `(1+1)/2 , (3+9)/2` ) = (1,6)

F1F2 = 2ae = 6

`rArr` ae = 3

But e = `1/2`       `:.` a = 6

b2 = a2(1-e2) = 36 ( 1-`1/4` ) = 27

Thus required equation is

`((x-1)^2)/(27) + ((y-6)^2)/36 = 1

I am planning to write more post on Equation of a Hyperbola and factoring polynomials of degree 3. Keep checking my blog.

Some more Examples to Find the General Equation of an Ellipse:

Ex 3: Find the equation of the ellipse  given that the centre is (4,-1) and one of its focus is (1,-1) and passing through (8,0)

Sol : From the given data since the major axis is parallel to the x axis , the equation is of the form

`((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) =1`

The centre is C(h,k) is (4,-1).

example 3 to find general equation

`((x-4)^2)/(a^2) + ((y+1)^2)/(b^2) =1`

It passes through (8,0) `:. 16/(a^2) + 1/(b^2) = 1`

But CF1 = ae = 3

b2= a2(1-e2) = a2 - a2e2 = a2-9

(1) `rArr 16/(a^2) + 1/(a^2 -9) =1`

`rArr ` 16a2 - 144+a2 = a4 - 9a2

`rArr` a4 - 26a2 +144= 0

`rArr` a2 = 18 or a2 = 8

Case (i) a2 = 18 `rArr` b2 = a2 -9 = 18-9 = 9

Case (ii) a2 = 8  `rArr` b2 = 8-9 = -1 which is not possible

`:. ` a2 = 18 and b2 = 9

Thus the equation is `((x-4)^2)/18 + ((y+1)^2)/9 = 1`

Trig Problem Solver

Greek Mathematician Ptolemy, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. Trigonometry was mainly concerned with establishing the relations between sides and angles of a triangle. The trigonometry consists of angles, quadrants, ratios and Identities, Compound Angles, and trigonometrical Equations

Understanding trig function is always challenging for me but thanks to all math help websites to help me out.

Trig problem solver  can be used to  find the angle values without using the tables.

We have addition formlae Sin(A+B) = SinA Cos B + Cos A Sin B

Cos(A+B) = CosACos B - SinA Sin B

and    Tan (A+B) =  tanA + tan B

1 - TanA.Tan B
Solved Problems on Trigonometry:
Let us use trig problem solver  to  do this problem
Ex 1: Without using tables, find sin 75º
Sol:-
We use the first formula Sin(A + B) = sin A cos B + cos A sin B
Sin 75º = sin(45º + 30º)= sin 45º cos 30º + cos 45º sin 30º
=  1 . √3  +  1 . 1    =  √3 + 1  = 1.732 +1  =  2.732 =  0.97
√2   2      √2  2           2√2      2x1.414      2.818
Ans is 0.97
Checking with the tables we find sin 75º =  0.9659
Ex 2: Without using tables find cos 75º
Sol: Let us use the formula  Cos (A+B) = cos A cosB - sinA sinB
cos 75º= Cos(45º +  30º)= cos 45º cos30 -  sin 45º sin30º
=   1 . √3  -  1 .1 =   √3 -  1 =  1.732 - 1  =   0.732   =   0.26
√2   2    √2.2        2√2        2 x 1.414     2.818
Ans cos 75º= 0.26
Checking with the tables we  find sin 75º = 0.2588
Ex 3: Find   tan 90 º without using tables
Sol: Tan (A+B) =  tan A + tan B
1 - tan A tanB
Tan 90 = tan (60º + 30º) =  tan 60º + tan 30º  =  √3 +  1/√3 =   √3 + 1 /√3  = √3 + 1 / √3 =    infinity=  ∞
1 - tan 60ºtan 30º    1 - √3.(1)         1  -    1                 0
√3
Hence we get tan 90º = infinity.
More Problems on Trigonometry:
Ex 5:If a = 4 units  and angle A = 45° find the circum-radius.
Sol:Here we use the formula  'a' = 2 R sin A where 'a' is the length of the side opposite angle A and R is circum- radius.
Hence 4 =  2R sin 45º =  2R x 1 / √2
4  ÷ 1 / √2 = 2R
4√2 =  2 R  =>  2√2 = R
Ans : Circum- radius of the circle with  A = sin45º and 'a' = 4 units is 2√2 units.
Ex 5: In any triangle prove that a(sinB - sin C) + b(sinC - sin A ) +c (sinA - sinB ) = 0
Ans: a= 2R sin A and b = 2R sin B and c = 2R sin C
Hence 2R sinA ( sinB - sinC) + 2R sin B( sin C - sin A ) + 2R sin C ( sin A - sin B)
=  2R sinA sinB - 2R sinA sin C + 2R sin B sinC - 2R sinB sin A + 2 R sinC sinA - 2 R sinCsinB
= 0 since( +2RsinAsinB cancels - 2R sinA sinB )and ( + 2RsinB sin C cancels - 2R sinB sinc) and (+2RsinC sin A cancels - 2R SinCsinA)

Ans for the above problem is zero and we have proved it by using trig problem solver.
Ex 6:Prove that Σ a sin (B − C) = 0

Sol:

Σ a sin (B−C) = a sin(B − C) + b sin(C − A) + c sin (A − B)

= 2R sinA sin(B−C) + 2R sinB sin(C − A) + 2R sinC sin(A − B)

sinA = sin(B + C), sinB = sin(C + A) ; sinC = sin(A + B)

By using the above formula, we get

= 2R sin(B + C) sin(B − C) + 2R sin(C + A) sin(C − A) + 2R sin(A +B) sin(A − B)

= 2R [sin2B − sin2C + sin2C − sin2A + sin2A − sin2B]

After simplifying this, we get

= 0

Ex 7: Solve: 2sin2x + sin22x = 2

Sol:

2 sin2x + sin22x = 2

sin22x = 2 − 2sin2x

= 2(1 − sin2x)

sin22x = 2 cos2x

⇒ 4sin2x cos2x − 2 cos2x = 0

⇒ 2(1 − cos2x) cos2x − cos2x = 0

Simplify the above terms, we get

⇒ 2cos4x − cos2x = 0

⇒ cos2x (2 cos2x − 1) = 0

⇒ cos2x = 0 , cos2x =`(1)/(2)` = `(1/sqrt(2))^2`

After simplifying this, we get

⇒ cos2x = cos2`pi/2`

⇒ x = nπ ±`pi/2` , n ∈ Z

Ex 8: Solve : √3 sin x + cosx = 2

Sol:

This is of the form a cosx + b sinx = c,

So dividing the equation by √ ( √3)2 + 12  ) or 2

We get

√3/2 sin x +1/2 cosx = 1 ⇒ sinπ/3 . sinx + cosπ/3 . cos x = 1

cos( x - π/3) = 1

cos(x - π/3) = cos 0

x - π/3 = 2nπ ± 0

x = 2nπ + π/3; n ∈ Z

Ex 3: In a triangle ABC prove that a sinA − b sinB = c sin(A − B)

Sol:

By sine formulae we have

a/sinA =b/sinB = c/sinC = 2R

a = 2R sinA, b = 2R sinB, c = 2R sin C

a sinA − b sinB = 2RsinA sinA − 2R sinB sinB

= 2R (sin2A − sin2B)

= 2R sin(A + B) sin(A − B)

= 2R sin(180 − C) sin (A − B)

= 2R sinC sin(A − B)

= c sin (A − B)
Practice Problems Using Trig Problem Solver:

Pro 1: Solve the following:

(i) sin3x = sinx

(ii) sin 4x + sin2x = 0

Ans: (i) nπ ±`pi/4` , nπ (ii) `(npi)/3` , `[(2n + 1)pi]/2`

Pro 2: Solve the problem: sin2x + sin6x + sin4x = 0

Ans: x = nπ ± `pi/3` ; n ∈ Z

Pro 3:  Prove the following: cos−1 (2x2 − 1) = 2cos−1x

Pro 4:  Find the solution: sin[ cos-11/2]

Ans: √3/2