1
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
The number of normals that can be drawn through the point $(9,6)$ to the parabola $y^2=4 x$ is
A
0
B
1
C
2
D
3
2
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $(2,3)$ is the focus and $x-y+3=0$ is the directrix of a parabola, then the equation of the tangent drawn at the vertex of the parabola is
A
$x-y-2=0$
B
$x-y+2=0$
C
$x-y+5=0$
D
$x-y-5=0$
3
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The equation of the common tangent to the parabola $y^2=8 x$ and the circle $x^2+y^2=2$ is $a x+b y+2=0$. If $-\frac{a}{b}>0$, then $3 a^2+2 b+1=$
A
5
B
4
C
3
D
2
4
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

    Consider the parabola $25\left[(x-2)^2+(y+5)^2\right]=(3 x+4 y-1)^2$, match the characteristic of this parabola given in List I with its corresponding item in List II.

    $$ \begin{array}{lll} \hline & \text { List I } & \text { List II } \\\\ \hline \text { I } & \text { Vertex } & \text { (A) } 8 \\\\ \hline \text { II } & \text { length of latus rectum } & \text { (B) }\left(\frac{29}{10}, \frac{-38}{10}\right) \\\\ \hline \text { III } & \text { Directrix } & \text { (C) } 3 x+4 y-1=0 \\\\ \hline \text { IV } & \begin{array}{l} \text { One end of the latus } \\\\ \text { rectum } \end{array} & \text { (D) }\left(\frac{-2}{5}, \frac{-16}{5}\right) \\\\ \hline \end{array} $$

    The correct answer is

A
I-B, II-E, III-C, IV-D
B
I-D, II-A, III-C, IV-B
C
I-B, II-A, III-C, IV-D
D
I-D, II-B, III-C, IV-A
TS EAMCET Subjects
EXAM MAP