1
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
The combined equation of a possible pair of adjacent sides of a square with area 16 square units whose centre is the point of intresection of the lines $x+2 y-3=0$ and $2 x-y-1=0$ is
A
$(2 x-y-1+4 \sqrt{5})(x+2 y-3+4 \sqrt{5})=0$
B
$(2 x-y-1-4 \sqrt{5})(x+2 y-4 \sqrt{5})=0$
C
$(2 x-y-2 \sqrt{5})(x+2 y+2 \sqrt{5})=0$
D
$(2 x-y-1-2 \sqrt{5})(x+2 y-3+2 \sqrt{5})=0$
2
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
If the line $2 x+b y+5=0$ forms an equilateral to triangle with $a x^{2}-96 b x y+k y^{2}=0$, then $a+3 k=$
A
$3 b$
B
192
C
$4 b^{2}$
D
102
3
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If the distance from a variable point $P$ to the point $(4,3)$ is equal to the perpendicular distance from $P$ to the line $x+2 y-1=0$, then the equation of the locus of the point $P$ is
A
$4 x^2+4 x y+y^2-38 x+26 y+124=0$
B
$4 x^2-4 x y+y^2-38 x-26 y+124=0$
C
$4 x^2-4 x y+y^2+38 x+26 y+124=0$
D
$4 x^2-4 x y+y^2-38 x+26 y+124=0$
4
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$(0, k)$ is the point to which the origin is to be shifted by the translation of the axes so as to remove the first degree terms from the equation $a x^2-2 x y+b y^2-2 x+4 y+1=0$ and $\frac{1}{2} \tan ^{-1}(2)$ is the angle through which the coordinate axes are to be rotated about the origin to remove the $x y$-term from the given equation, then $a+b=$
A
1
B
-2
C
3
D
-4
TS EAMCET Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12