Let $e$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $a=5, b=4$ and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is $lx+my=27$,then $l+m=$

In a group of $100$ persons,$75$ speak English and $40$ speak Hindi. Each person speaks at least one of the two languages. If the number of persons who speak only English is $\alpha$ and the number of persons who speak only Hindi is $\beta$,then the eccentricity of the ellipse $25(\beta^2 x^2 + \alpha^2 y^2) = \alpha^2 \beta^2$ is $.......$

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the coordinate axes is

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

Let $P$ be a parabola with vertex $(2,3)$ and directrix $2x+y=6$. Let an ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ of eccentricity $e=\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$ is: