What conic section is represented by the polar equation r = 1 / 4 - 6cos theta

Question

asked Mar 21, 2020 147k views

2 Answers

Zhivko Draganov · Answer 1 · 2020-03-26T10:51:01+0000

Answer:

Option 2 - Hyperbola

Explanation:

Given : The polar equation
$r=(1)/(4-6\cos\theta)$

To find : What conic section is represented by the polar equation?

Solution :

To find the conic section first we convert the polar into Cartesian equation

We know,
r=√(x^2+y^2) and
$x=r\cos\theta$

$r=(1)/(4-6\cos\theta)$

$4r-6r\cos\theta=1$

Substitute the value of r,

4(√(x^2+y^2))-6x=1

4√(x^2+y^2)=1+6x

Squaring both side,

16(x^2+y^2)=(1+6x)^2

16x^2+16y^2=1+36x^2+12x

16y^2=20x^2+12x+1

Applying completing the square we get,

16y^2=20(x+(3)/(10))^2-(4)/(5)

16y^2-20(x+(3)/(10))^2=-(4)/(5)

$(16y^2)/(-(4)/(5))-{20(x+(3)/(10))^2}{-(4)/(5)}=1$

$-(y^2)/((1)/(4))+{(x+(3)/(10))^2}{(1)/(25)}=1$

${(x+(3)/(10))^2}{(1)/(25)}-(y^2)/((1)/(4))=1$

This is in the form of hyperbola equation i.e.
(x^2)/(a^2)-(y^2)/(b^2) =1

Therefore, The given conic section is a hyperbola.

Hence, Option 2 is correct.