Mathematics

Question

Solve the equation for an ellipse for y. Assume that y > 0. y^2/a^2 + x^2/b^2 = 1

2 Answer

  • multiply each term by a^2b^2:-

    b^2y^2 + a^2x^2 = a^2b^2
    subtract a^2x^2 from both sides
    b^2y^2  = a^2b^2  - a^2x^2
    Now divide both sides  by b^2

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

    take positive square root ( because y > 0)

    y  = a sqrt(1 - x^2/b^2)
  • Answer:

    The required value of y is:

    [tex]y=a\sqrt{1-\dfrac{x^2}{b^2}}[/tex]

    Step-by-step explanation:

    We have to solve the equation for an ellipse for y.

    That means we have to find the value of y in terms of x from the given equation.

    The equation of an ellipse is given as:

    [tex]\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1[/tex]

    We will multiply both side by [tex]a^2b^2[/tex] to obtain:

    [tex]b^2y^2+a^2x^2=a^2b^2[/tex]

    Now we will take the term of variable 'x' to the right hand side to obtain:

    [tex]b^2y^2=a^2b^2-a^2x^2\\\\y^2=\dfrac{a^2b^2-a^2x^2}{b^2}\\\\y^2=\dfrac{a^2b^2}{b^2}-\dfrac{a^2x^2}{b^2}\\\\y^2=a^2-\dfrac{a^2x^2}{b^2}\\\\y^2=a^2(1-\dfrac{x^2}{b^2})[/tex]

    No on taking square root on both the side we obtain:

    [tex]y=a\sqrt{1-\dfrac{x^2}{b^2}}[/tex]

    Hence, the required value of y is:

    [tex]y=a\sqrt{1-\dfrac{x^2}{b^2}}[/tex]

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