Let $F_1$ & $F_2$ be the foci of an ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ such that a ray from $F_1$ strikes the elliptical mirror at the point $P$ and get reflected. Then equation of angle bisector of the angle between incident ray and reflected ray can be 

The line, $ lx + my + n = 0$  will cut the ellipse $\frac{{{x^2}}}{{{a^2}}}$ $+$ $\frac{{{y^2}}}{{{b^2}}}$ $= 1 $ in points whose eccentric angles differ by $\pi /2$  if :

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.

$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are

$(A)$ $(3,0)$ and $(0,2)$

$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is

$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$

$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$

$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is

$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$

$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$

$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$

$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$

 Give the answer question $1,2$ and $3.$

If $\alpha $ and $\beta $ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

Let a tangent to the Curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $A B$ is $………$