For the parabola y^2+6y-2x+5=0,match the item | vedclass

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

Question

(B) Given parabola: $y^2+6y-2x+5=0$Completing the square for $y$:$y^2+6y+9 = 2x-5+9$$(y+3)^2 = 2(x+2)$Comparing with $(y-k)^2 = 4a(x-h)$,we get:Vertex

Vedclass · Accepted Answer

$F, A, C, E$

Vedclass · Answer

(B) Given parabola: $y^2+6y-2x+5=0$Completing the square for $y$:$y^2+6y+9 = 2x-5+9$$(y+3)^2 = 2(x+2)$Comparing with $(y-k)^2 = 4a(x-h)$,we get:Vertex $(h, k) = (-2, -3)$. This matches $(F)$.$4a = 2 \Rightarrow a = \frac{1}{2}$.Focus $(h+a, k) = (-2+\frac{1}{2}, -3) = (-\frac{3}{2}, -3)$. This matches $(A)$.Equation of directrix: $x = h-a$ $\Rightarrow x = -2-\frac{1}{2}$ $\Rightarrow x = -\frac{5}{2}$ $\Rightarrow 2x+5=0$. This matches $(C)$.Equation of axis: $y = k \Rightarrow y+3=0$. This matches $(E)$.Thus,the correct matching is $(I-F, II-A, III-C, IV-E)$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

Similar Questions

If $x-y-3=0$ is a normal drawn through the point $(5,2)$ to the parabola $y^2=4x$,then the slope of the other normal that can be drawn through the same point to the parabola $y^2=4x$ is

If the tangents of the parabola $y^2=8x$ passing through the point $P(1,3)$ touch the parabola at $A$ and $B$,then the area (in sq. units) of $\triangle PAB$ is

The equation of the parabola whose vertex and focus are $(0, 4)$ and $(0, 2)$ respectively,is

If the line $x = my + k$ touches the parabola $x^2 = 4ay$,then $k = $

The axis of a parabola is parallel to the $Y$-axis. If this parabola passes through the points $(1,0), (0,2), (-1,-1)$ and its equation is $ax^2 + bx + cy + d = 0$,then $\frac{ad}{bc} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module