If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25 x^{2}+4 y^{2}=1$ to the parabola $y^{2}=4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}$ equals
$7982$
$2898$
$2929$
$3289$
Let $E$ be the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ and $C$ be the circle ${x^2} + {y^2} = 9$. Let $P$ and $Q$ be the points $(1, 2)$ and $(2, 1)$ respectively. Then
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$ | $List-II$ |
If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
Let $\mathrm{E}$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4) .$ If $m x-y=4, m\,>\,0$ is a tangent to the ellipse $\mathrm{E}$, then the value of $5 \mathrm{~m}^{2}$ is equal to $.....$
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to