Let the end points M and N of the diameter 4x – y = 15 of the circle  $x^2+y^2-6x+6y-16=0$ are also on the tangents at the end points of the major axis of the ellipse respectively, such that MN is also tangent to the same ellipse. The major axis is along the line y = x then which of the following option(s) is (are) TRUE?

 $M(3+\sqrt{2},-3+4\sqrt{2})N(3-\sqrt{2},-3-4\sqrt{2})$
End points of major axis are  $A(\frac{-5}{\sqrt{2}},\frac{-5}{\sqrt{2}})A\text{'}(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$
length of major axis  $=2a=AA\text{'}=10$
 $\boxed{a=5}$
centre $=\frac{A+A\text{'}}{2}=(0,0)$ 
focii of ellipse are S $(\alpha ,\alpha )$and  $S\text{'}(-\alpha ,-\alpha )$ $x+y=5\sqrt{2},x+y=-5\sqrt{2}\&4x-y=15$  are tangents to the ellipse
 $\left(SP\right)(S\text{'}Q)=b^2$
 $\Rightarrow \left|\frac{5\sqrt{2}-2\alpha }{\sqrt{2}}\right|\left|\frac{5\sqrt{2}+2\alpha }{\sqrt{2}}\right|=\left|\frac{15-3\alpha }{\sqrt{17}}\right|\left|\frac{15+3\alpha }{\sqrt{17}}\right|=b^2\Rightarrow {\alpha }^2=8$
 $$\begin{gathered} S(2\sqrt{2}, 2\sqrt{2}), S\text{'}(-2\sqrt{2},-2\sqrt{2}) \\ S{S}^{\text{'}}=2ae=\left(4\sqrt{2}\right)\sqrt{2}=8 \\ \Rightarrow ae=4 \\ \Rightarrow e=\frac{4}{5} \end{gathered}$$
length of latus rectum  $=\frac{2b^2}{a}=\frac{2\left(17\right)}{5}=\frac{34}{5}$
equation of directrices : $x+y=\frac{25\sqrt{2}}{4} or \frac{-25\sqrt{2}}{4}$