Question

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

Answer 1

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 2

Answer:

The required value of y is:

$y=a\sqrt{1-\dfrac{x^2}{b^2}}$

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:

$\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1$

We will multiply both side by $a^2b^2$ to obtain:

$b^2y^2+a^2x^2=a^2b^2$

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

$b^2y^2=a^2b^2-a^2x^2\\\\y^2=\dfrac{a^2b^2-a^2x^2}{b^2}\\\\y^2=\dfrac{a^2b^2}{b^2}-\dfrac{a^2x^2}{b^2}\\\\y^2=a^2-\dfrac{a^2x^2}{b^2}\\\\y^2=a^2(1-\dfrac{x^2}{b^2})$

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

$y=a\sqrt{1-\dfrac{x^2}{b^2}}$

Hence, the required value of y is:

$y=a\sqrt{1-\dfrac{x^2}{b^2}}$