No, her answer is not reasonable because one yard is equal to 3 feet.
<h3>Answer:</h3>
Equation of the ellipse = 3x² + 5y² = 32
<h3>Step-by-step explanation:</h3>
<h2>Given:</h2>
- The centre of the ellipse is at the origin and the X axis is the major axis
- It passes through the points (-3, 1) and (2, -2)
<h2>To Find:</h2>
- The equation of the ellipse
<h2>Solution:</h2>
The equation of an ellipse is given by,

Given that the ellipse passes through the point (-3, 1)
Hence,

Cross multiplying we get,
- 9b² + a² = 1 ²× a²b²
- a²b² = 9b² + a²
Multiply by 4 on both sides,
- 4a²b² = 36b² + 4a²------(1)
Also by given the ellipse passes through the point (2, -2)
Substituting this,

Cross multiply,
- 4b² + 4a² = 1 × a²b²
- a²b² = 4b² + 4a²-------(2)
Subtracting equations 2 and 1,
- 3a²b² = 32b²
- 3a² = 32
- a² = 32/3----(3)
Substituting in 2,
- 32/3 × b² = 4b² + 4 × 32/3
- 32/3 b² = 4b² + 128/3
- 32/3 b² = (12b² + 128)/3
- 32b² = 12b² + 128
- 20b² = 128
- b² = 128/20 = 32/5
Substituting the values in the equation for ellipse,


Multiplying whole equation by 32 we get,
3x² + 5y² = 32
<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
<h3>Explanation:</h3>
1. PQ║TS, PQ ≅ TS, PT and QS are transversals to the parallel lines . . . given
2. ∠P ≅ ∠T . . . alternate interior angles at PT
3. ∠Q ≅ ∠S . . . alternate interior angles at QS
4. ΔPQR ≅ ΔTSR . . . ASA postulate
_____
You can use any pair of angles together with the sides PQ and TS. If you use the vertical angles and one of ∠T or ∠S, then you must invoke the AAS postulate for congruence, as the side is not between the two angles.
You can take it apart. There are a top and bottom (both the same) right triangle. So you can find the area of that by multiplying 8*6 and divide by two. Then multiply by two because there are 2 triangles.
You are left with three rectangular sides: One 10x10, one 10x6, and one 10x8.
So your whole equation looks like this: A = 2[(8*6)/2]+(10*10)+(10*6)+(10*8)