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Write the equation in standard form for the ellipse x2+9y22x26=0x^2 + 9y^2 - 2x - 26 = 0.

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Q. Write the equation in standard form for the ellipse x2+9y22x26=0x^2 + 9y^2 - 2x - 26 = 0.
  1. Complete the Square for xx: Now, we complete the square for the xx terms. To do this, we take half of the coefficient of xx, which is 1-1, square it, and add it to both sides.\newlinex22x+12+9y2=26+1x^2 - 2x + 1^2 + 9y^2 = 26 + 1\newlinex22x+1+9y2=27x^2 - 2x + 1 + 9y^2 = 27
  2. Rewrite Equation for x: We don't need to complete the square for the yy terms because there is no linear yy term. Now, we can rewrite the equation to show the perfect square trinomial for xx.(x1)2+9y2=27(x - 1)^2 + 9y^2 = 27
  3. Divide by 2727: Next, we divide the entire equation by 2727 to get the standard form of the ellipse equation. \newline(x1)2/27+9y2/27=27/27(x - 1)^2/27 + 9y^2/27 = 27/27
  4. Simplify the Equation: Simplify the equation by reducing the fractions. (x1)227+y23=1\frac{(x - 1)^2}{27} + \frac{y^2}{3} = 1

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