Question

Let the ellipse \[ E=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] have eccentricity equal to the greatest value of the function \[ f(t)=-\frac34+2t-t^2 \] and the length of its latus rectum is $30$. Find the value of $a^2+b^2$.

Hint:

For ellipse problems, always connect eccentricity and latus rectum formulas carefully.

Answer 1

Correct Answer: 496

Step 1: Find the greatest value of the given function

f(t) = −3/4 + 2t − t²

This is a downward opening parabola. The maximum value occurs at

t = −b / 2a = −2 / (2 × −1) = 1

Maximum value:

f(1) = −3/4 + 2 − 1 = 1/4

Thus, eccentricity of the ellipse,
e = 1/4

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Step 2: Use eccentricity relation of ellipse

For the ellipse:

e² = 1 − (b²/a²)

(1/4)² = 1 − (b²/a²)

1/16 = 1 − (b²/a²)

b²/a² = 15/16

b² = (15/16)a²

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Step 3: Use formula for length of latus rectum

Length of latus rectum = 2b²/a

Given:

2b²/a = 30

Substitute b² = (15/16)a²

2 × (15/16)a² / a = 30

(15/8)a = 30

a = 16

a² = 256

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Step 4: Find b²

b² = (15/16) × 256

b² = 240

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Final Answer:

a² + b² = 256 + 240 = 496