A point $P$ moves so that distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is
When the coordinate axes are rotated through an angle $\theta$ in anti clockwise direction, if the transformed equation of $x^2+y^2+2 x y+2 x+6 y+1=0$ is $(2+\sqrt{3}) X^2+2 X Y+(2-\sqrt{3}) Y^2+a X+b Y+2=0$, then $3 a-b=$
If the lines $3 x+y-4=0, x-a y-10=0, b x+2 y+9=0$ form three successive sides of a rectangle in that order and the fourth side passes through $(1,2)$, then the area of that rectangle (in sq. units) is
The points $A(2,1), B(3,-2)$ and $C(a, b)$ are vertices of the rectangle $A B C D$. If the point $P(3,4)$ lies on $C D$ produced, then $5 a+10 b=$
TS EAMCET Papers
All year-wise previous year question papers