If $P \equiv (x, y)$,$F_1 \equiv (3, 0)$,$F_2 \equiv (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2$ equals

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the midpoint of the line segment $PQ$,then the locus of $M$ intersects the latus rectums of the given ellipse at the points

The eccentricity of an ellipse having centre at the origin,axes along the coordinate axes and passing through the points $(4, -1)$ and $(-2, 2)$ is

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 $A$ and $B$.
$1.$ The coordinates of $A$ and $B$ 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 $PAB$ 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 $P$ and the line $AB$ 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 for questions $1, 2$ and $3.$

If the lines $2x - y + 3 = 0$ and $4x + ky + 3 = 0$ are conjugate with respect to the ellipse $5x^2 + 6y^2 - 15 = 0$,then $k$ equals:

The locus of the midpoints of the intercepted portion of the tangents by the coordinate axes,which are drawn to the ellipse $x^2+2y^2=2$,is