If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:

Let $T_1$ and $T_2$ be two distinct common tangents to the ellipse $E: \frac{x^2}{6}+\frac{y^2}{3}=1$ and the parabola $P: y^2=12 x$. Suppose that the tangent $T_1$ touches $P$ and $E$ at the point $A_1$ and $A_2$, respectively and the tangent $T_2$ touches $P$ and $E$ at the points $A_4$ and $A_3$, respectively. Then which of the following statements is(are) true?

($A$) The area of the quadrilateral $A_1 A _2  A _3 A _4$ is $35$ square units

($B$) The area of the quadrilateral $A_1 A_2 A_3 A_4$ is $36$ square units

($C$) The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-3,0)$

($D$) The tangents $T_1$ and $T_2$ meet the $x$-axis at the point $(-6,0)$

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.

$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are

$(A)$ $(3,0)$ and $(0,2)$

$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is

$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$

$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$

$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is

$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$

$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$

$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$

$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$

 Give the answer question $1,2$ and $3.$

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

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

Suppose that the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ are $\left(f_1, 0\right)$ and $\left(f_2, 0\right)$ where $f_1>0$ and $f_2<0$. Let $P _1$ and $P _2$ be two parabolas with a common vertex at $(0,0)$ and with foci at $\left(f_1, 0\right)$ and $\left(2 f_2, 0\right)$, respectively. Let $T_1$ be a tangent to $P_1$ which passes through $\left(2 f_2, 0\right)$ and $T_2$ be a tangent to $P_2$ which passes through $\left(f_1, 0\right)$. The $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$, then the value of $\left(\frac{1}{m^2}+m_2^2\right)$ is