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german
3 years ago
10

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

Mathematics
2 answers:
In-s [12.5K]3 years ago
4 0
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)
DaniilM [7]3 years ago
4 0

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}}

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