The point of intersection of the tangent at 2 points on the ellipse whose ecentricity differs by 90 degree , how can we write this stmt?

Your question is not complete complete the complete question is, 
Find the locus of the point of intersection of tangents at the ends of  semi-conjugate diameters of an ellipse.
or
The eccentric angles of two points P and Q on the ellipse differ by $\frac{\mathrm{\pi }}{2}$ . Find the locus of point of intersection of tangents at 
P and Q.
For this let me explain you what is eccentric angles of extremities of two conjugate semi-diameters is, 


$$\begin{gathered} \mathrm{First} \mathrm{lets} \mathrm{know} \mathrm{about} \mathrm{Conjugate} \mathrm{dimaters}-\mathrm{two} \mathrm{diameters} \mathrm{y} = {\mathrm{m}}_1\mathrm{x} \mathrm{and} \mathrm{y} = {\mathrm{m}}_2\mathrm{x} \mathrm{of} \mathrm{an} \mathrm{ellipse} \mathrm{are} \mathrm{said} \mathrm{to} \mathrm{be} \mathrm{conjugate} \mathrm{if} \\ \mathrm{each} \mathrm{bisects} \mathrm{chords} \mathrm{parallel} \mathrm{to} \mathrm{each} \mathrm{other} \\ \mathrm{And} \mathrm{required} \mathrm{condition} \mathrm{for} \mathrm{any} \mathrm{two} \mathrm{diameters} \mathrm{to} \mathrm{be} \mathrm{conjugate} \mathrm{is} =\frac{-{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ \mathrm{Proof}-\mathrm{Let} \mathrm{y} = {\mathrm{m}}_2\mathrm{x} \mathrm{bisects} \mathrm{all} \mathrm{chords} \mathrm{of} \mathrm{the} \mathrm{ellipse} \mathrm{which} \mathrm{are} \mathrm{parallel} \mathrm{to} \mathrm{the} \mathrm{diameter} \mathrm{y} = {\mathrm{m}}_1\mathrm{x}. \mathrm{If} \left(\mathrm{h}, \mathrm{k} \right) \mathrm{be} \mathrm{mid} \mathrm{points} \\ \mathrm{of} \mathrm{the} \mathrm{chords} \mathrm{parallel} \mathrm{to} \mathrm{y} = {\mathrm{m}}_1\mathrm{x} \mathrm{then} \mathrm{it} \mathrm{will} \mathrm{lie} \mathrm{on} \mathrm{y} = {\mathrm{m}}_2\mathrm{x} \\ \mathrm{or} \mathrm{k} = {\mathrm{m}}_2\mathrm{h}.........\left(1\right) \\ \mathrm{But} \mathrm{if} \left(\mathrm{h},\mathrm{k} \right)\mathrm{be} \mathrm{the} \mathrm{mid} \mathrm{points} \mathrm{of} \mathrm{the} \mathrm{chord} , \mathrm{then} \mathrm{its} \mathrm{equation} \mathrm{is}, \\ \frac{\mathrm{hx}}{{\mathrm{a}}^2}+\frac{\mathrm{ky}}{{\mathrm{b}}^2}=\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2} \\ \mathrm{Its} \mathrm{slop} \mathrm{is} = \frac{-{\mathrm{hb}}^2}{{\mathrm{ka}}^2} \\ \mathrm{But} \mathrm{this} \mathrm{chord} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{y} = {\mathrm{m}}_1\mathrm{x} \\ \mathrm{So} \frac{-{\mathrm{hb}}^2}{{\mathrm{ka}}^2}= {\mathrm{m}}_1........\left(2\right) \\ \mathrm{But} \mathrm{from} \mathrm{equation} 1, \\ \frac{\mathrm{h}}{\mathrm{k}}=\frac{1}{{\mathrm{m}}_2} \\ \mathrm{or} \frac{-1}{{\mathrm{m}}_2}\times \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}= {\mathrm{m}}_1 \left[\mathrm{from} \mathrm{equation} 1\right] \\ \mathrm{or} {\mathrm{m}}_1\times {\mathrm{m}}_2=-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ \mathrm{And} \mathrm{it} \mathrm{is} \mathrm{required} \mathrm{condition} \mathrm{for} \mathrm{any} \mathrm{two} \mathrm{diameters} \mathrm{to} \mathrm{be} \mathrm{conjugate} \end{gathered}$$

Now Eccentric angles of extremities of two conjugate semi diameter-




$$\begin{gathered} \mathrm{Let} \mathrm{\theta } \mathrm{and} \mathrm{\varphi } \mathrm{be} \mathrm{the} \mathrm{eccentric} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{extremities} \mathrm{P} \mathrm{and} \mathrm{D} \mathrm{of} \mathrm{two} \mathrm{cinjugate} \\ \mathrm{semi}-\mathrm{diameters} \mathrm{CP} \mathrm{and} \mathrm{CD}. \\ \mathrm{Hence} \mathrm{P} \mathrm{is} \left(\mathrm{acos\theta }, \mathrm{bsin\theta }\right), \mathrm{D}\left(\mathrm{acos\varphi }, \mathrm{bsin\varphi }\right) \\ \mathrm{now} \mathrm{slop} \mathrm{of} \mathrm{CP}={\mathrm{m}}_1=\frac{\mathrm{bsin\theta }-0}{\mathrm{acos\theta }-0}=\frac{\mathrm{bsin\theta }}{\mathrm{acos\theta }} \\ \mathrm{now} \mathrm{slop} \mathrm{of} \mathrm{CD}={\mathrm{m}}_2=\frac{\mathrm{bsin\varphi }-0}{\mathrm{acos\varphi }-0}=\frac{\mathrm{bsin\varphi }}{\mathrm{acos\varphi }} \\ \mathrm{Since} \mathrm{CP} \mathrm{and} \mathrm{CD} \mathrm{are} \mathrm{conjugate} \mathrm{semi}-\mathrm{diameters}. \\ \mathrm{So} {\mathrm{m}}_1\times {\mathrm{m}}_2=\frac{-{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ \mathrm{or} \frac{{\mathrm{b}}^2\mathrm{sin\theta sin\varphi }}{{\mathrm{a}}^2\mathrm{cos\theta cos\varphi }}=\frac{-{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ \mathrm{or} \mathrm{cos\theta cos\varphi }+\mathrm{sin\theta sin\varphi }=0 \\ \mathrm{or} \cos \left(\mathrm{\varphi }-\mathrm{\theta }\right)=0 \\ \left(\mathrm{\varphi }-\mathrm{\theta }\right)=\frac{\mathrm{\pi }}{2} \mathrm{or} \mathrm{\varphi }=\frac{\mathrm{\pi }}{2}+\mathrm{\theta } \\ \mathrm{hence} \mathrm{P} \mathrm{is} \left(\mathrm{acos\theta }, \mathrm{bsin\theta }\right) \mathrm{then} \mathrm{D} \mathrm{is} \left\{\mathrm{acos}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right), \mathrm{b} \sin \left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)\right\} \\ \mathrm{or} \left(-\mathrm{asin\theta }, \mathrm{bcos\theta }\right). \end{gathered}$$

Now coming to the question, 

$$\begin{gathered} \mathrm{The} \mathrm{tangent} \mathrm{at} \mathrm{P} \mathrm{is} \frac{\mathrm{xcos\theta }}{\mathrm{a}}+\frac{\mathrm{ysin\theta }}{\mathrm{b}}=1 \\ \mathrm{Tangent} \mathrm{at} \mathrm{D} \mathrm{is} \frac{-\mathrm{xsin\theta }}{\mathrm{a}}+\frac{\mathrm{ycos\theta }}{\mathrm{b}}=1 \\ \mathrm{In} \mathrm{order} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{locus} \mathrm{of} \mathrm{the} \mathrm{points} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{these} \mathrm{tangents} \mathrm{we} \\ \mathrm{have} \mathrm{to} \mathrm{eliminate} \mathrm{\theta } \mathrm{for} \mathrm{which} \mathrm{we} \mathrm{square} \mathrm{and} \mathrm{add}. \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}\times 1+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}\times 1+0=2 \\ \mathrm{or} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=2 \end{gathered}$$