Let the length of the latus rectum of an ellipse (x^2)/(a^2)+(y^2)/(b^2)=1,(ab), be 30 .

Question

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be 30 . If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2 t-t^2$, then $\left(a^2+b^2\right)$ is equal to

Accepted Answer

length of latus rectum of ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,(a>b)$ is given as:

$l=\frac{2{b}^{2}}{a}=30\Rightarrow {b}^{2}=15a$

It is given that eccentricity is the maximum value of the function $f(t)=-{t}^{2}+2t-\frac{3}{4}$

max value of quadratic function $a{x}^{2}+bx+c$ is given as $\frac{-D}{4a}=\frac{4ac-{b}^{2}}{4a}$

$f(t{)}_{max}=\frac{4(-1)(-\frac{3}{4})-(2{)}^{2}}{4(-1)}=\frac{3-4}{-4}=\frac{1}{4}$

so eccentricity $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}=\frac{1}{4}$

squaring

$1-\frac{{b}^{2}}{{a}^{2}}=\frac{1}{16}$

put value of ${b}^{2}=15a$

$\begin{gathered} \frac{{a}^{2}-15a}{{a}^{2}}=\frac{1}{16} \ 16{a}^{2}-240a={a}^{2} \ 15{a}^{2}-240a=0 \ a(15a-240)=0 \ a
e 0,a=\frac{240}{15}=16 \end{gathered}$

Now, value of ${a}^{2}+{b}^{2}={a}^{2}+15a{{b}^{2}=15a}$

$\begin{gathered} =(16{)}^{2}+15	imes 16 \ =256+240=496 \end{gathered}$

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