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 $a^2+b^2$.

• A. 496
• B. 250
• C. 376
• D. 175

Hint:

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

Answer 1

The Correct Option is A

To solve this problem, we need to use the properties of ellipses and the given function to determine the values of \(a^2\) and \(b^2\). Here is the step-by-step solution:

1. The standard form of the ellipse given is:  
   \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
2. The eccentricity \(e\) of the ellipse is given by: 
   \(e = \sqrt{1-\frac{b^2}{a^2}}\)
3. We are told the eccentricity \(e\) is equal to the greatest value of the function: 
   \(f(t) = -\frac{3}{4} + 2t - t^2\)
4. Find the maximum value of \(f(t)\). To do this, compute the derivative \(f'(t)\) and find its critical points: 
   \(f'(t) = 2 - 2t\) 
   Setting \(f'(t) = 0\), we have: 
   \(2 - 2t = 0 \implies t = 1\)
5. Calculate \(f(1)\): 
   \(f(1) = -\frac{3}{4} + 2(1) - 1^2 = \frac{1}{4}\) 
   Thus, the maximum value of the function is \(\frac{1}{4}\).
6. Therefore, the eccentricity \(e\) of the ellipse is \(\frac{1}{4}\): 
   \(\sqrt{1-\frac{b^2}{a^2}} = \frac{1}{4}\)
7. Next, the length of the latus rectum of the ellipse is given by: 
   \(\frac{2b^2}{a} = 30\)
8. We now have two equations:
   • \(1 - \frac{b^2}{a^2} = \left(\frac{1}{4}\right)^2 = \frac{1}{16}\)
   • \(\frac{b^2}{a^2} = \frac{15}{16} \quad \text{(Rearranging the first equation)}\)
   • \(\frac{2b^2}{a} = 30\) implies \(b^2 = 15a\)
9. From \(b^2 = 15a\), substitute into \(\frac{b^2}{a^2} = \frac{15}{16}\): 
   \(\frac{15a}{a^2} = \frac{15}{16} \Rightarrow a = 16\)
10. Substitute back \(a = 16\) into \(b^2 = 15a\): 
    \(b^2 = 15 \times 16 = 240\)
11. Find \(a^2 + b^2\): 
    \(a^2 = 16^2 = 256\) 
    \(a^2 + b^2 = 256 + 240 = 496\)
12. Therefore, the correct answer is: 
    496