Find the equation of the ellipse with the major axis along the $x-$axis and passing through the points $(4, 3)$ and $(-1, 4)$.

$A$ tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersects the coordinate axes at $A$ and $B$. Then the locus of the circumcentre of triangle $AOB$ (where $O$ is the origin) is:

Let $A=\{(\alpha, \beta) \in R \times R :|\alpha-1| \leq 4 \text{ and }|\beta-5| \leq 6\}$ and $B=\left\{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}$. Then

If the normal drawn at the point $P(\frac{\pi}{4})$ on the ellipse $x^2+4y^2-4=0$ meets the ellipse again at $Q(\alpha, \beta)$,then $\alpha=$

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 $A$ and $B$.
$1.$ The coordinates of $A$ and $B$ 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 $PAB$ 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 $P$ and the line $AB$ 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 for questions $1, 2$ and $3.$

If $x \cos \alpha + y \sin \alpha = 4$ is a tangent to $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$,then the value of $\alpha$ is