(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

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

(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 $(I-F, II-A, III-C, IV-E)$.

Class 11 Mathematics 2 Conic Sections: Parabola, Ellipse and Hyperbola Parabola

Easy

For the parabola $y^2+6y-2x+5=0$,match the items in List-$I$ with the suitable item in List-$II$ given below:
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 $(D)$ $2x+y+3=0$
$(E)$ $y+3=0$
$(F)$ $(-2, -3)$

AP EAMCET 2018 All AP EAMCET

A
$F, A, E, C$
B
$F, A, C, E$
C
$A, B, C, D$
D
$F, A, C, D$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

2 Conic Sections: Parabola, Ellipse and Hyperbola

Subtopic

Parabola

Previous Year

AP EAMCET 2018 Paper All AP EAMCET years →

The line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = x + 2$ at the points $P$ and $Q$. If the coordinates of the point $X$ are $(\sqrt{3}, 0)$, then the value of $XP \cdot XQ$ is

MediumWBJEE 2025

View Solution

If $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $\lambda$ is equal to

DifficultAP EAMCET 2008

View Solution

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

View Solution

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 20y$ to the ends of its latus rectum is

MediumMHT CET 2025

View Solution

Consider the circle $C: x^2+y^2=4$ and the parabola $P: y^2=8x$. If the set of all values of $\alpha$,for which three chords of the circle $C$ on three distinct lines passing through the point $(\alpha, 0)$ are bisected by the parabola $P$,is the interval $(p, q)$,then $(2q-p)^2$ is equal to.............

DifficultJEE MAIN 2024

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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

The line $y - \sqrt{3}x + 3 = 0$ cuts the parabola $y^2 = x + 2$ at the points $P$ and $Q$. If the coordinates of the point $X$ are $(\sqrt{3}, 0)$, then the value of $XP \cdot XQ$ is

If $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $\lambda$ is equal to

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 20y$ to the ends of its latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module