In an ellipse  $\frac{x^2}{10}+\frac{y^2}{6}=1$. If focal chords  $PSP\text{'}$ and  $QSQ\text{'}$ are at right angles to each other, then  $\frac{1-e^2}{\left(SP\right)(SP\text{'})}+\frac{1-e^2}{\left(SQ\right)(SQ\text{'})}=\frac{N}{15}$. The greatest integer less than or equal to N is

 

Length of focal chord
If $PP\text{'}$  is inclined at angle $\text{'}{\theta }^{\text{'}}$  then  $QQ\text{'}$ is inclined at angle  $\frac{\pi }{2}+\theta$
 $QQ\text{'}=\frac{2a{b}^2}{a^2{\cos }^2\theta +b^2{\sin }^2\theta }$
 $\frac{1}{SP}+\frac{1}{S\text{'}P\text{'}}=\frac{2}{l}\Rightarrow \frac{1-e^2}{\left(SP\right)(SP\text{'})}=\frac{2}{a(PP\text{'})}$
 $\frac{1}{SQ}+\frac{1}{SQ\text{'}}=\frac{2}{l}\Rightarrow \frac{1-e^2}{\left(SQ\right)(SQ\text{'})}=\frac{2}{a(QQ\text{'})}$
 $=\frac{16}{60}=\frac{4}{15}$
 $N=4$