Solve the formula $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { f ^ { 2 } } = 1 ; y$
for the indicated variable.
A)$y=\sqrt{f(2-x a)(1+x a)}, y=-\sqrt{f(2-x a)(1+x a)}$
B)$y=f \sqrt{\left(1-\frac{x}{a}\right)\left(1+\frac{x}{a}\right)}, y=-f \sqrt{\left(1-\frac{x}{a}\right)\left(1+\frac{x}{a}\right)}$
C)$y=f \sqrt{\left(a-\frac{x}{a}\right)\left(a+\frac{x}{a}\right)}, y=-f \sqrt{\left(a-\frac{x}{a}\right)\left(a+\frac{x}{a}\right)}$

D)$y=\sqrt{f\left(1-\frac{x}{a}\right)^{2}}, y=-\sqrt{f\left(1-\frac{x}{a}\right)^{2}}$

E)$Y=f(1-x a)(1+x a), Y=-\sqrt{f(1-x a)(1+x a)}$

Question

Solve the formula $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { f ^ { 2 } } = 1 ; y$
for the indicated variable.
A)$y=\sqrt{f(2-x a)(1+x a)}, y=-\sqrt{f(2-x a)(1+x a)}$ 
B)$y=f \sqrt{\left(1-\frac{x}{a}\right)\left(1+\frac{x}{a}\right)}, y=-f \sqrt{\left(1-\frac{x}{a}\right)\left(1+\frac{x}{a}\right)}$ 
C)$y=f \sqrt{\left(a-\frac{x}{a}\right)\left(a+\frac{x}{a}\right)}, y=-f \sqrt{\left(a-\frac{x}{a}\right)\left(a+\frac{x}{a}\right)}$

D)$y=\sqrt{f\left(1-\frac{x}{a}\right)^{2}}, y=-\sqrt{f\left(1-\frac{x}{a}\right)^{2}}$

E)$Y=f(1-x a)(1+x a), Y=-\sqrt{f(1-x a)(1+x a)}$

Quizplus · Accepted Answer

To solve for $y$, we isolate $y$ in the equation $\frac{x^2}{a^2} + \frac{y^2}{f^2} = 1$. Rearranging, we get $\frac{y^2}{f^2} = 1 - \frac{x^2}{a^2}$. Multiplying both sides by $f^2$, we have $y^2 = f^2\left(1 - \frac{x^2}{a^2}\right)$. Factoring the right side, we get $y^2 = f^2\left(\frac{a^2 - x^2}{a^2}\right) = f^2\left(1 - \frac{x}{a}\right)\left(1 + \frac{x}{a}\right)$. Taking the square root of both sides gives $y = \pm f \sqrt{\left(1 - \frac{x}{a}\right)\left(1 + \frac{x}{a}\right)}$, which matches choice B.

Solve the Formula x2a2+y2f2=1;y\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { f ^ { 2 } } = 1 ; ya2x2​+f2y2​=1;y

Solve the Formula $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { f ^ { 2 } } = 1 ; y$