Solve the equation for an ellipse for y. Assume that y > 0. y^2/a^2 + x^2/b^2 = 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:
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:
\(y=a\sqrt{1-\dfrac{x^2}{b^2}}\)