Ellipse practice · Mixed difficulty

An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=1$ as its semi-minor axis and diameter of the circle $x^2+(y-2)^2=4$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is

If $B$ and $B'$ are the ends of minor axis and $S$ and $S'$ are the foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1,$ then area of the rhombus $SBS'B'$ will be

If the length of the major axis of an ellipse is three times the length of its minor axis, its eccentricity, is

The eccentricity of the ellipse $9 x^2+5 y^2-30 y=0$ is

The condition on $a$ and $b$ for which two distinct chords of the ellipse $\frac{x^2}{2 a^2}+\frac{y^2}{2 b^2}=1$ passing through $(a,-b)$ are bisected by the line $x+y=b$, is

If the latusrectum of an ellipse is equal to one half of its minor axis, then the eccentricity is equal to

If PSQ is a focal chord of the ellipst $16 x^2+25 y^2=400$ such that $S P=8$, then $S Q=$

The equation of an ellipse whose eccentricity is $\frac{1}{2}$ and the vertices are, $\left( {4,0} \right)$ and $\left( {10,0} \right)$ is

Equation of the ellipse whose foci are (2,2) and (4,2) and the major axis is of length 10 is

Let the line $y = mx$ and the ellipse $2{x^2} + {y^2} = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left( { - \frac{1}{{3\sqrt 2 }},0} \right)$ and $\left( {0,\,\beta } \right),$ then $\beta$ is equal to

If the chord joining points $P\left( \alpha \right)$ and $Q\left( \beta \right)$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ subtends a right angle at the vertex $A\left( {a,0} \right),$ then $\tan \alpha /2\tan \beta /2 =$

Statement I : An equation of a common tangent to the parabola $y^2=16 \sqrt{3} x$ and the ellipse $2 x^2+y^2=4$ is $y=2 x+2 \sqrt{3}$
Statement II : If the line $y=m x+\frac{4 \sqrt{3}}{m},(m \neq 0)$ is a common tangent to the parabola $y^2=16 \sqrt{3} x$ and the ellipse $2 x^2+y^2=4$, then $m$ satisfies $m^2+2 m^2=24$

The eccentricity of the conic ${x^2} - 4\;x + 4{y^2} = 12$ is

The coordinates of the foci of the ellipse $3 x^2+4 y^2-12 x$ $-8 y+4=0$ are

$S(3,4)$ and $S^{\prime}(9,12)$ are two focii of an ellipse. If the foot of the perpendicular from $S$ on a tangent to the ellipse has the coordinates $(1,-4)$, then the eccentricity of the ellipse, is

The circle $x^2+ y^2 = c^2$ contains the ellipse $x^2/a^2 + y^2/b^2 = 1$ if

If the normal at one end of the latusrectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the one end of the minor axis, then

If the tangent drawn at point $\left(t^2, 2 t\right)$ on the parabola $y^2=4 x$ is same as the normal drawn at point $(\sqrt{5} \cos \theta, 2 \sin \theta)$ on the ellipse $4 x^2+5 y^2=20$. Then, the values of $t$ and $\theta$ are

Number of points on the ellipse $\frac{x^2}{50}+\frac{y^2}{20}=1$ from which pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

The equation of the ellipse whose foci are at $\left( { \pm \,2,0} \right)$ and eccentricity is $\frac{1}{2}$, is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, Then,

The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, which make complementary angles with $x$-axis, is

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $S_1$ and $S_2$. If $A$ be the area of $\triangle P S_1 S_2$, then the maximum value of $A$, is

The equation of the ellipse with eccentricity $1 / 2$ and foci at $( \pm 1,0)$ is

The locus of poles of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with respect to concentric ellipse $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$, is

If from a point $P$ tangents $P Q$ and $P R$ are drawn to the ellipse $\frac{x^2}{2}+y^2=1$, so that equation of $Q R$ is $x+3 y=1$, then coordinates of $P$ is

The locus of the point of intersection of the perpendicular tangents to ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is

The equation of an ellipse, whose eccentricity is $1 / 2$ and the vertices are $(4,0)$ and $(10,0)$ is

The locus of mid – points of the line segments joining (-3, -5) and the points on the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ is

Suppose $x$ and $y$ are real numbers and that $x^2+9 y^2-4 x+6 y+4=0$, then the maximum value of $(4 x-9 y)$.

The foci of the conic section $25 x^2+16 y^2-150 x=175$ are