The point (2,1) is translated parallel to the line L:x-y=4 by 2√3 units. If the new point Q lies in the third quadrant, then the equation of the line…

Let the image of parabola $x^{2}=4 y$, in the line $x-y=1$ be $(y+a)^{2}=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to

Let the domain of the function $f(x)=\log _{3} \log _{5} \log _{7}\left(9 x-x^{2}-13\right)$ be the interval $(\mathrm{m}, \mathrm{n})$. Let the hyperbola $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 \mathrm{~m}}{3}$. Then $\mathrm{b}^{2}-\mathrm{a}^{2}$ is equal to :

The distance between the points (3, 4) and (6, 8) is:

If P is a point on the circle $x^{2}+y^{2}=4, \mathrm{Q}$ is a point on the straight line $5 x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is $\_\_\_\_$.

Let a point $A$ lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :