Question

Let $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1\left(a>b\right)$ be a given ellipse, length of whose latus rectum is $10$. If its eccentricity is the maximum value of the function, $\Phi \left(t\right)=\left(\frac{5}{12}\right)+t-t^2$, then $a^2+b^2$ is equal to:

• A. $135$
• B. $116$
• C. $126$
• D. $145$

Answer 1

The correct option is C

$126$

Explanation for the correct answer:

Step-1: Finding the relation between $a$ and $b$

The given equation of the ellipse is: $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1\left(a>b\right)$.

Now, we know that the length latus rectum of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$is given by $l=\frac{2b^2}{a}$.

Given, that the length of the latus rectum is $10$. So, we must get:

$$\begin{gathered} l=\frac{2b^2}{a} \\ \Rightarrow \frac{2b^2}{a}=10 \\ \Rightarrow \frac{b^2}{a}=5 \end{gathered}$$

Step-2: finding the maximum value of the function $\Phi \left(t\right)$

$$\begin{gathered} \Phi \left(t\right)=\left(\frac{5}{12}\right)+t-t^2 \\ =\left(\frac{5}{12}\right)-\left(t^2-2.t.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4} \\ =\left(\frac{5}{12}+\frac{1}{4}\right)-{\left(t-\frac{1}{2}\right)}^2 \\ =\left(\frac{5+3}{12}\right)-{\left(t-\frac{1}{2}\right)}^2 \\ =\frac{2}{3}-{\left(t-\frac{1}{2}\right)}^2 \end{gathered}$$

Now, ${\left(t-\frac{1}{2}\right)}^2\ge 0$, $\forall t$. So, the maximum value of $\Phi \left(t\right)$ is given by: $\Phi {\left(t\right)}_{max}=\frac{2}{3}$.

Step-3: Finding another relation between $a$ and $b$.

We know that the eccentricity of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$, is given by: $e=\sqrt{1-\frac{b^2}{a^2}}$.

By the question, the eccentricity of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$is the maximum value of the function $\Phi \left(t\right)$ i.e. $\Phi {\left(t\right)}_{max}=\frac{2}{3}$.

So, we get:

$$\begin{gathered} e=\frac{2}{3} \\ \Rightarrow \sqrt{1-\frac{b^2}{a^2}}=\frac{2}{3} \\ \Rightarrow 1-\frac{b^2}{a^2}={\left(\frac{2}{3}\right)}^2 \\ \Rightarrow \frac{b^2}{a^2}=1-\frac{4}{9} \\ \Rightarrow \frac{b^2}{a^2}=\frac{5}{9} \end{gathered}$$

Step-4: Finding the values of $a$and $b$

We obtain two equations:

$\frac{b^2}{a}=5$ $\left(1\right)$

and $\frac{b^2}{a^2}=\frac{5}{9}$. $\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$, we get:

$$\begin{gathered} \frac{{\displaystyle \frac{b^2}{a}}}{{\displaystyle \frac{b^2}{a^2}}}=\frac{5}{{\displaystyle \frac{5}{9}}} \\ \Rightarrow a=9 \end{gathered}$$

Now, from $\left(1\right)$, we get:

$$\begin{gathered} b^2=5a \\ =5\times 9 \\ =45 \end{gathered}$$

Step-5: Finding the value of $a^2+b^2$

Now, we calculate the value of $a^2+b^2$, using the values obtained in Step-4 as follows:

$$\begin{gathered} {\mathrm{a}}^2+{\mathrm{b}}^2=9^2+45 \\ \Rightarrow a^2+b^2=81+45 \\ \Rightarrow a^2+b^2=126 \end{gathered}$$

Therefore, the correct answer is option (C).