The value of $a$ for which the image of the point $(a, a - 1)$ with respect to the line mirror $3x + y = 6a$ is the point $(a^2 + 1, a)$ is:

Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$. Then,the foot of the perpendicular from the origin on the variable line $\frac{x}{a} + \frac{y}{b} = 1$ lies on

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a $\triangle ABC$ are $x-y+5=0$ and $x+2y+5=0$,respectively. If $A$ is $(1, -2)$,then the equation of the straight line $BC$ is

If the reflection of a point $A(2,3)$ in the $X$-axis is $B$; the reflection of $B$ in the line $x+y=0$ is $C$,and the reflection of $C$ in $x-y=0$ is $D$,then the point of intersection of the lines $CD$ and $AB$ is:

The equation of a straight line which passes through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and is perpendicular to $x \sec \theta + y \operatorname{cosec} \theta = a$ is

Find the equation of a line which passes through $(2 \cos^3 \theta, 2 \sin^3 \theta)$ and is perpendicular to the line $x \cos \theta - y \sin \theta = 2 \cos 2 \theta$.