To solve this problem, we need to find the value of \((a^2+b^2)\) for the given ellipse and function constraints.
First, recall the formula for the length of the latus rectum of an ellipse. For the ellipse given by \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) with \(a > b\), the length of the latus rectum is \(\dfrac{2b^2}{a}\).
We are told that the length of the latus rectum is 30, so we have:
\(\dfrac{2b^2}{a}=30\)
From this, we obtain:
\(b^2 = 15a\)
Next, find the maximum value of the function \(f(t)=-\dfrac{3}{4}+2t-t^2\). This is a quadratic function of the form \(f(t) = -t^2 + 2t - \dfrac{3}{4}\), which is concave down.
The maximum value of a quadratic function \(ax^2 + bx + c\) when \(a < 0\) is at \(x = -\dfrac{b}{2a}\).
Substituting the values, \(t = -\dfrac{2}{2(-1)} = 1\) is the vertex.
Substituting \(t = 1\) back into \(f(t)\) gives:
\(f(1) = -1^2 + 2\cdot1 - \dfrac{3}{4} = 1 - \dfrac{3}{4} = \dfrac{1}{4}\)
The eccentricity \(e\) of the ellipse is equal to the maximum value of the function, \(e = \dfrac{1}{4}\).
For an ellipse, the eccentricity \(e\) is given by:
\(e = \sqrt{1-\dfrac{b^2}{a^2}}\)
Substituting \(e = \dfrac{1}{4}\) and \(b^2 = 15a\), we have:
\(\dfrac{1}{4} = \sqrt{1-\dfrac{15a}{a^2}}\), which simplifies to \(\dfrac{1}{4} = \sqrt{1-\dfrac{15}{a}}\)
Squaring both sides results in:
\(\dfrac{1}{16}=1-\dfrac{15}{a}\)
This simplifies to \(\dfrac{15}{a}=1-\dfrac{1}{16}\) or \(\dfrac{15}{a}=\dfrac{15}{16}\)
Thus, \(a=16\).
Now use \(b^2 = 15a\):
\(b^2 = 15 \cdot 16 = 240\)
Finally, calculate \(a^2 + b^2\):
\(a^2 = 16^2 = 256\)
\(a^2 + b^2 = 256 + 240 = 496\)
Therefore, the value of \((a^2+b^2)\) is 496.
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$