The centre of the ellipse$\frac{{{{(x + y - 2)}^2}}}{9} + \frac{{{{(x - y)}^2}}}{{16}} = 1$ is

Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1<0, y_2<0$, be the end points of the latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are

$(A)$ $x^2+2 \sqrt{3} y=3+\sqrt{3}$

$(B)$ $x^2-2 \sqrt{3} y=3+\sqrt{3}$

$(C)$ $x^2+2 \sqrt{3} y=3-\sqrt{3}$

$(D)$ $x^2-2 \sqrt{3} y=3-\sqrt{3}$

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( {\frac{1}{2},\;1} \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle ${x^2} + {y^2} = 1$ and the hyperbola ${x^2} - {y^2} = 1$. The equation of the ellipse in the standard form, is

An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is

The equation of normal at the point $(0, 3)$ of the ellipse $9{x^2} + 5{y^2} = 45$ is

Let $L$ be a common tangent line to the curves $4 x^{2}+9 y^{2}=36$ and $(2 x)^{2}+(2 y)^{2}=31$. Then the square of the slope of the line $L$ is ..... .