Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is

The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at

The equation of the normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $(a\cos \theta ,\;b\sin \theta )$ is

Find the equation for the ellipse that satisfies the given conditions: Centre at $(0,\,0),$ major axis on the $y-$ axis and passes through the points $(3,\,2)$ and $(1,\,6)$

An ellipse, with foci at $(0, 2)$ and $(0, -2)$ and minor axis of length $4$, passes through which of the following points?

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 ($2$)