Let F₁, F₂ be two foci of the ellipse and PT and PN be the tangent and the normal respectively to the ellipse at point P. Then

Normal at a point on the ellipse bisects the angle subtended by the foci at that point. So, Option (A) is correct.

Now, let $\angle F_{1}P{F}_2$=2x

Then, $\angle F_{2}PN$=x

Since, PN is perpendicular to PT,   $\angle F_{2}PT$=(90∘−x)

Lets extend $F_{1}L$.  Supplement of ∠F₁PF₂ is (180∘−2x) 

Since, $\angle F_{2}PT$=$\frac{1}{2}$(${180}^0$−∠F₁PF₂ ) ,PT bisects the angle (${180}^0$-∠F₁PF₂ )

So, Option (C) is also correct