The eccentricity of the ellipse $ (x - 3)^2 + (y - 4)^2 =$ $\frac{{{y^2}}}{9}\,$  is

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 $3 x+4 y=12 \sqrt{2}$ is a tangent to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{9}=1$ for some a $\in \mathrm{R},$ then the distance between the foci of the ellipse is

Let the line $2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0, \mathrm{k}>0$, intersect the $\mathrm{x}$-axis and $\mathrm{y}$-axis at the points $\mathrm{A}$ and $\mathrm{B}$, respectively. If the equation of the circle having the line segment $\mathrm{AB}$ as a diameter is $\mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}-2 \mathrm{y}=0$ and the length of the latus rectum of the ellipse $\mathrm{x}^2+9 \mathrm{y}^2=\mathrm{k}^2$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime, then $2 \mathrm{~m}+\mathrm{n}$ is equal to

Slope of common tangents of parabola $(x -1)^2 = 4(y -2)$ and ellipse ${\left( {x - 1} \right)^2} + \frac{{{{\left( {y - 2} \right)}^2}}}{2} = 1$ are $m_1$ and $m_2$ ,then $m_1^2 + m_2^2$ is equal to

If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to