For the parabola y^2+6y-2x=-5,consider the fo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $y^2+6y-2x=-5$Completing the square for $y$:$y^2+6y+9 = 2x-5+9$$(y+3)^2 = 2x+4$$(y+3)^2 = 2(x+2)$Comparing this with the standard

Vedclass · Accepted Answer

$I$ is true,$II$ is false

Vedclass · Answer

(B) Given equation: $y^2+6y-2x=-5$Completing the square for $y$:$y^2+6y+9 = 2x-5+9$$(y+3)^2 = 2x+4$$(y+3)^2 = 2(x+2)$Comparing this with the standard form $(y-k)^2 = 4a(x-h)$,we get:Vertex $(h, k) = (-2, -3)$. Thus,statement $I$ is true.Here,$4a = 2$,so $a = \frac{1}{2}$.The directrix of the parabola $(y-k)^2 = 4a(x-h)$ is given by $x = h-a$.$x = -2 - \frac{1}{2} = -\frac{5}{2}$$2x = -5 \Rightarrow 2x+5 = 0$.Statement $II$ says the directrix is $y+3=0$,which is false.Therefore,$I$ is true and $II$ is false.

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?

Similar Questions

If $(2 t^2, 4 t)$ is a point on the parabola $y^2 = 8x$ such that its focal distance is $3$,then $t =$

The line $y = 6x + 1$ touches the parabola $y^2 = 24x$. The coordinates of a point $P$ on this line,from which the tangent to $y^2 = 24x$ is perpendicular to the line $y = 6x + 1$,is

Let $P$ be the point $(1, 0)$ and $Q$ be a point on the locus $y^2 = 8x$. The locus of the midpoint of $PQ$ is

The equation of the line touching both parabolas $y^2=4x$ and $x^2=-32y$ is

The equation of the tangent to the parabola $y^2=12x$,which makes an angle $30^{\circ}$ with the positive direction of the $X$-axis is given by $x-\sqrt{3}y+9=0$. The point of contact is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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?