what is hkabc of the ellipse with the formula x^2 + 4y^2 - 2x -16y + 1 = 0

Question

please help!

Answer 1

Given equation x^2+4y^2-2x-16y+1 = 0

Add 16 to each side.

x^2+4y^2-2x-16y+1+16 = 0+16

x^2-2x+1+4y^2-16y+16 = 16

x^2-2x+1+4(y^2-4y+4) = 16

(x-1)^2+4(y-2)^2 =16

Divide to each side by 16.

(x-1)^2/16+4(y-2)^2/16 = 16/16

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

Compare it to standard form of ellipse is (x-h)^2/a^2+(y-k)^2/b^2 = 1

Where a > b.

(h,k) = (1,2)

a = 4, b = 2.

c = √(a^2-b^2)

c = √(16-4)

c = √8

Comments

The value of  .

Answer 2

The ellipse equation

To change the expressions (x ²- 2x ) and (y ² - 4y ) 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 16 to set it equal to 1.

Compare it to standard form of ellipse

a ² > b ²

If the larger denominator is under the "x" term, then the ellipse is horizontal.

center (h, k ) = (1, 2)

a  = length of semi-major axis = 4

b  = length of semi-minor axis = 2.

c is the distance from the center to each focus.

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