Let $P$ be a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with foci ${F_1}$ and ${F_2}$. If $A$ is the area of the triangle $P{F_1}{F_2}$, then maximum value of $A$ is

If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$, then $12\; m ^{2}$ is equal to

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 ($2$)

If $a$ and $c$ are positive real numbers and the ellipse $\frac{{{x^2}}}{{4{c^2}}} + \frac{{{y^2}}}{{{c^2}}} = 1$ has four distinct points in common with the circle $x^2 + y^2 = 9a^2$ , then

The locus of a variable point whose distance from $(-2, 0)$ is $\frac{2}{3}$ times its distance from the line $x = - \frac{9}{2}$, is

The position of the point $(1, 3)$ with respect to the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$