Math  /  Geometry

QuestionIn Exercises 3 and 4 , solve for yy in terms of xx.
3. y29+x24=1\frac{y^{2}}{9}+\frac{x^{2}}{4}=1
4. x236+y225=1\frac{x^{2}}{36}+\frac{y^{2}}{25}=1

Studdy Solution

STEP 1

1. We are given two equations representing ellipses.
2. We need to solve each equation for y y in terms of x x .

STEP 2

1. Solve the first equation for y y in terms of x x .
2. Solve the second equation for y y in terms of x x .

STEP 3

Start with the first equation:
y29+x24=1 \frac{y^{2}}{9} + \frac{x^{2}}{4} = 1
Isolate the y2 y^2 term:
y29=1x24 \frac{y^{2}}{9} = 1 - \frac{x^{2}}{4}

STEP 4

Multiply both sides by 9 to solve for y2 y^2 :
y2=9(1x24) y^{2} = 9 \left(1 - \frac{x^{2}}{4}\right)

STEP 5

Simplify the expression:
y2=99x24 y^{2} = 9 - \frac{9x^{2}}{4}

STEP 6

Take the square root of both sides to solve for y y :
y=±99x24 y = \pm \sqrt{9 - \frac{9x^{2}}{4}}

STEP 7

Start with the second equation:
x236+y225=1 \frac{x^{2}}{36} + \frac{y^{2}}{25} = 1
Isolate the y2 y^2 term:
y225=1x236 \frac{y^{2}}{25} = 1 - \frac{x^{2}}{36}

STEP 8

Multiply both sides by 25 to solve for y2 y^2 :
y2=25(1x236) y^{2} = 25 \left(1 - \frac{x^{2}}{36}\right)

STEP 9

Simplify the expression:
y2=2525x236 y^{2} = 25 - \frac{25x^{2}}{36}

STEP 10

Take the square root of both sides to solve for y y :
y=±2525x236 y = \pm \sqrt{25 - \frac{25x^{2}}{36}}
The solutions are:
For equation 3: y=±99x24 y = \pm \sqrt{9 - \frac{9x^{2}}{4}}
For equation 4: y=±2525x236 y = \pm \sqrt{25 - \frac{25x^{2}}{36}}

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