For the ellipse 25x^2+9y^2-126y+200x= 59,

Question

find the standard form. then find the center, vertices, foci, and?

Answer 1

The  given  ellipse equation is 25x²+9y²-126y +200x = 59

25x²+200x +9y²-126y =59

Add  400 and 441 to each side

25x²+200x +400 +9y²-126y +441 = 59+400+441

25 ( x²+8x+16 ) +9 ( y²-14y +49 ) = 900

25 ( x+4)²+9 ( y - 7)²= 900

[ 25 ( x+4 )²/ 900 ] +[ 9 ( y - 7)²/ 900 ]  = 0

( x+4 )²/ 36 +( y - 7)²/ 100  = 1

The ellipse  standard  formula  is   ( x - (-4)²) / 36 + ( y - 7)²/ 100 = 1

Therefore centre ( h,k ) = ( -4, 7 )

The  vertical  major  axis  standard  formula  is ( x-h)²/ b²+( y-k)²/ a²= 0

( x -( -4 ))²/ 100 +( y - 7)²/ 36  = 1

a²= 100   =>  a = 10

b² = 36  => b = 6

Therefore  vertices (  k ± a , h ) =  ( 17 , -4 ) , ( -3,-4 )

c²  = a²- b²

c² = 64  => c = 8

Therefore  foci  = ( k ± c, h ) = ( 15 , -4 ) , ( -1, -4 ).

Comments

The ellipse 25x ²+ 9y ²- 126y  + 200x  = 59

Vertices are : (h , k + a ), (h , k - a ) = (-4, 17), (-4, -3) and

Foci: (h , k + c ), (h, k - c ) =  (-4, 15), (-4, -1).

Answer 2

The ellipse equation

To change the expressions (x ²+ 8x ) and (y ² - 14y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 900 to set it equal to 1.

Compare the standardform of ellipse

a ² > b ²

If the larger denominator is under the "y " term, then the ellipse is vertical.Center (h, k ).

a  = length of semi-major axis, b  = length of semi-minor axis.

a  = 10, b  = 6

Center ( - 4, 7 )

Vertices: (h , k + a ), (h , k - a )

= ( -4, 7+10 ), ( -4, 7-10 )

Vertices  (- 4, 17 ), (- 4, - 3 )

c  is the distance from the center to each focus.

Foci: (h , k + c ), (h, k - c )

= (-4, 7+8), (-4, 7-8)

Foci ( - 4, 15 ), ( - 4, - 1 ).