Let a line $L$ pass through the point of intersection of the lines $b x+10 y-8=0$ and $2 x-3 y=0$, $b \in R -\left\{\frac{4}{3}\right\}$. If the line $L$ also passes through the point $(1,1)$ and touches the circle $17\left( x ^{2}+ y ^{2}\right)=16$, then the eccentricity of the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{b^{2}}=1$ 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 ($2$)

The eccentricity of ellipse $(x-3)^2 + (y -4)^2 = \frac{y^2}{9} +16 ,$ is –

If $\frac{{\sqrt 3 }}{a}x + \frac{1}{b}y = 2$ touches the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ then its, eccentric angle $\theta $ is equal to: ……………. $^o$