For the parabola $y^2+6y-2x=-5$,consider the following statements:
$I$. The vertex is $(-2, -3)$.
$II$. The directrix is $y+3=0$.
Which of the following is correct?
$I$. The vertex is $(-2, -3)$.
$II$. The directrix is $y+3=0$.
Which of the following is correct?
- ABoth $I$ and $II$ are correct
- B$I$ is true,$II$ is false
- CBoth $I$ and $II$ are false
- D$I$ is false,$II$ is true
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Similar Questions
Three normals are drawn from the point $(3, 0)$ to the parabola $y^2 = 4x$,meeting the parabola at points $P, Q,$ and $R$. Match the following:
Column-$I$ Column-$II$ $(A)$ Circumradius of $\Delta PQR$ $(P)$ $5/2$ $(B)$ Area of $\Delta PQR$ $(Q)$ $(5/2, 0)$ $(C)$ Centroid of $\Delta PQR$ $(R)$ $(2/3, 0)$ $(D)$ Circumcenter of $\Delta PQR$ $(S)$ $2$
| Column-$I$ | Column-$II$ |
|---|---|
| $(A)$ Circumradius of $\Delta PQR$ | $(P)$ $5/2$ |
| $(B)$ Area of $\Delta PQR$ | $(Q)$ $(5/2, 0)$ |
| $(C)$ Centroid of $\Delta PQR$ | $(R)$ $(2/3, 0)$ |
| $(D)$ Circumcenter of $\Delta PQR$ | $(S)$ $2$ |
Medium
View SolutionIf a normal chord of a parabola $y^2 = 4ax$ subtends a right angle at the origin,then the slope of that normal chord is
$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is
DifficultTS EAMCET 2013
View Solution$TP$ and $TQ$ are tangents of the parabola $y^2 = 8x$ at $P$ and $Q$ respectively. If the chord $PQ$ passes through the point $(-2, 3)$ and the locus of point $T$ is $y = mx + c$,then $(m + c)$ is equal to -
The equation of the given curve is $x^2-4x+4y-8=0$. Match the following:
List-$I$ List-$II$ $(A)$ Focus $(I)$ $(4,2)$ $(B)$ Vertex $(II)$ $(3,2)$ $(C)$ One end of the latus rectum $(III)$ $(2,3)$ $(D)$ Point of intersection of the axis and directrix $(IV)$ $(2,4)$ $(V)$ $(2,2)$
The correct matching is:
| List-$I$ | List-$II$ |
| $(A)$ Focus | $(I)$ $(4,2)$ |
| $(B)$ Vertex | $(II)$ $(3,2)$ |
| $(C)$ One end of the latus rectum | $(III)$ $(2,3)$ |
| $(D)$ Point of intersection of the axis and directrix | $(IV)$ $(2,4)$ |
| $(V)$ $(2,2)$ |
The correct matching is:
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