The normal at a variable point $P$ on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}= 1$  of eccentricity e meets the axes of the ellipse in $ Q$  and $R$  then the locus of the mid-point of $QR$  is a conic with an eccentricity $e' $  such that :

Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$

Let $F_1\left(x_1, 0\right)$ and $F_2\left(x_2, 0\right)$, for $x_1<0$ and $x_2>0$, be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.

($1$)The orthocentre of the triangle $F_1 M N$ is

($A$) $\left(-\frac{9}{10}, 0\right)$   ($B$) $\left(\frac{2}{3}, 0\right)$    ($C$) $\left(\frac{9}{10}, 0\right)$    ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$

($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is

($A$) $3: 4$     ($B$) $4: 5$     ($C$) $5: 8$     ($D$) $2: 3$

Givan the answer qestion ($1$) and ($2$)

Let $P(2,2)$ be a point on an ellipse whose foci are $(5,2)$ and $(2,6)$, then eccentricity of ellipse is 

The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x-$ axis at $Q$ and $R,$ respectively. Then the area(in sq. units) of the triangle $PQR$ is

A point on the ellipse, $4x^2 + 9y^2 = 36$, where the normal is parallel to the line, $4x -2y-5 = 0$ , is