The coordinate axes are rotated about the origin in the counter-clockwise direction through an angle $60^{\circ}$. If $a$ and $b$ are the intercepts made on the new axes by a straight line whose equation referred to the original axes is $x+y=1$,then $\frac{1}{a^2}+\frac{1}{b^2}=$

If the axes are translated to the orthocentre of the triangle formed by the points $A(7,5), B(-5,-7), C(7,-7)$,then the coordinates of the incentre of the triangle in the new system are

$A$ line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle $\theta$ keeping the origin fixed,this line $L$ has the intercepts $p$ and $q$. Then

$A$ vector $\vec{a}$ has components $3p$ and $1$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\vec{a}$ has components $p+1$ and $\sqrt{10}$,then a value of $p$ is equal to

The transformed equation $3x^2 + 3y^2 + 2xy = 2$,when the coordinate axes are rotated through an angle $45^{\circ}$,is

If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$,then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ is: