(B) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$. Multiplying by $8$,we get $\frac{8}{r} = 1 + 3 \cos \theta$. This is

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

(B) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$. Multiplying by $8$,we get $\frac{8}{r} = 1 + 3 \cos \theta$. This is in the standard polar form of a conic section $\frac{l}{r} = 1 + e \cos \theta$,where $e$ is the eccentricity. Comparing the two,we find $e = 3$. Since the eccentricity $e = 3 > 1$,the conic section represents a hyperbola.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Hyperbola

Difficult

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

A
$A$. $A$ rectangular hyperbola
B
$B$. $A$ hyperbola
C
$C$. An ellipse
D
$D$. $A$ parabola

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

10-2. Parabola, Ellipse, Hyperbola

Subtopic

Hyperbola

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.

Difficult

View Solution

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the coordinate axes at the distinct points $A$ and $B$,then the locus of the midpoint of $AB$ is

DifficultJEE MAIN 2018

View Solution

The locus of the midpoint of the system of parallel chords parallel to the line $y = 2x$ for the hyperbola $9x^2 - 4y^2 = 36$ is

DifficultWBJEE 2024

View Solution

The locus of the midpoints of the chord of the circle $x^{2}+y^{2}=25$ which is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ is

DifficultJEE MAIN 2021

View Solution

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}=576$.

Medium

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

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Similar Questions

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the coordinate axes at the distinct points $A$ and $B$,then the locus of the midpoint of $AB$ is

The locus of the midpoint of the system of parallel chords parallel to the line $y = 2x$ for the hyperbola $9x^2 - 4y^2 = 36$ is

The locus of the midpoints of the chord of the circle $x^{2}+y^{2}=25$ which is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ is

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}=576$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module