By shifting the origin to the point $(h, 5)$ by the translation of coordinate axes,if the equation $y=x^3-9x^2+cx-d$ transforms to $Y=X^3$,then $\left(d-\frac{c}{h}\right)=$

$(a, b)$ is the point to which the origin has to be shifted by translation of axes so as to remove the first-degree terms from the equation $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. If the angle by which the axes are to be rotated in the positive direction about the origin to remove the $xy$-term from the equation $ax^2 + 23abxy + by^2 = 0$ is $\theta$,then $\tan 2\theta =$

After the coordinate axes are rotated through an angle $\frac{\pi}{4}$ in the anti-clockwise direction without shifting the origin,if the equation $x^2+y^2-2x-4y-20=0$ transforms to $ax^2+2hxy+by^2+2gx+2fy+c=0$ in the new coordinate system,then $\left|\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right|=$

When the coordinate axes are rotated through an angle $\frac{\pi}{4}$ in the positive direction,an equation is transformed to $x^2+y^2-6x+8y+21=0$. Then the original equation is

The point $P(4,1)$ undergoes the following transformations in succession:
$(i)$ origin is shifted to the point $(1,6)$ by translation of axes
(ii) translation through a distance of $2$ units along the positive direction of $X$-axis
(iii) rotation of axes through an angle of $90^{\circ}$ in the positive direction
Then the coordinates of the point $P$ in its final position are

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 $\frac{1}{a^2}+\frac{1}{b^2}=$