Find the center, vertices and foci of the ellipse 9x^2 + 4y^2 + 36x – 8y + 4 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the equation 9x^2 + 4y^2 + 36x – 8y + 4 = 0

9x^2 + 4y^2 + 36x – 8y + 4 = 0

collect the x and y terms together

=> 9(x^2 + 4x) + 4(y^2 – 2y) = -4

complete the squares of the terms within brackets

=> 9( x^2 + 4x + 4) + 4(y^2 – 2y + 1) = -4 + 36 + 4

=> 9(x + 2) ^2 + 4(y – 1) ^2 = 36

divide by 36

=> (x + 2)^2 / 4 + (y – 1)^2 / 9 = 1

This is of the form (x – h)^2/ b^2 + ( y – k)^2/a^2 = 1

The center is given by (h, k ) = ( -2 , 1)

The vertices are (h , k + a) and (h, k – a) or ( -2 , 4) and (-2, -2)

The foci are (h, k + sqrt( a^2 – b^2)) and (h, k - sqrt( a^2 – b^2)) or ( -2 , 1 + sqrt 5) and ( -2, 1 – sqrt 5)

The required center is (-2, 1). The vertices are ( -2 , 4) and (-2, -2) and the foci are ( -2 , 1 + sqrt 5) and ( -2, 1 – sqrt 5)

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