The length of the focal chord of the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1$which is inclined to x – axis at an angle 45° is .........

$$\begin{gathered} Given ellipse \frac{x^2}{16}+\frac{y^2}{9}=1 \\ where a^2=16,b^2=9 and inclination of focal chord={45}^0 \\ Now length of focal chord=\frac{2a{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta } \\ =\frac{2\left(4\right)\left(9\right)}{16\left({\displaystyle \frac{1}{2}}\right)+9\left(\frac{1}{2}\right)} \\ =\frac{144}{25} \end{gathered}$$