Parabola · Mathematics · TS EAMCET
MCQ (Single Correct Answer)
1
$(1,1)$ is the vertex and $x+y+1=0$ is the directrix of a parabola. If $(a, b)$ is its focus and $(c, d)$ is the point of intersection of the directrix and the axis of the parabola, then $a+b+c+d=$
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2
The axis of a parabola is parallel to $Y$-axis. If this parabola passes through the points $(1,0),(0,2),(-1,-1)$ and its equation is $a x^{2}+b x+c y+d=0$, then $\frac{a d}{b c}=$
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3
$S=y^{2}-4 a x=0, S^{\prime}=y^{2}+a x=0$ are two parabolas and $P(t)$ is a point on the parabola $S^{\prime}=0$. If $A$ and $B$ are the feet of the perpendiculars from $P$ on to coordinate $2 x_{4}$ and $A B$ is a tangent to the parabola $S=0$ at the point $Q\left(t_{1}\right)$, then $t_{1}=$
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4
If the focal chord of the parabola $x^2=12 y$, drawn through the point $(3,0)$ intersects the parabola at the points $P$ and $Q$ then the sum of the reciprocals of the abscissae of the points $P$ and $Q$ is
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5
If the normal drawn at the point $P(9,9)$ on the parabola $y^2=9 x$ meets the parabola again at $Q(a, b)$, then $2 a+b=$
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6
$P$ and $Q$ are the extremities of a focal chord of the parabola $y^2=4 a x$. If $P=(9,9)$ and $Q=(p, q)$, then $p-q=$
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7
The number of normals that can be drawn through the point $(9,6)$ to the parabola $y^2=4 x$ is
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8
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
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9
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 $3 a^2+2 b+1=$
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10
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
TG EAPCET 2024 (Online) 9th May Morning Shift