Answer:
Option 2 - Hyperbola
Step-by-step explanation:
Given : The polar equation [tex]r=\frac{1}{4-6\cos\theta}[/tex]
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, [tex]r=\sqrt{x^2+y^2}[/tex] and [tex]x=r\cos\theta[/tex]
[tex]r=\frac{1}{4-6\cos\theta}[/tex]
[tex]4r-6r\cos\theta=1[/tex]
Substitute the value of r,
[tex]4(\sqrt{x^2+y^2})-6x=1[/tex]
[tex]4\sqrt{x^2+y^2}=1+6x[/tex]
Squaring both side,
[tex]16(x^2+y^2)=(1+6x)^2[/tex]
[tex]16x^2+16y^2=1+36x^2+12x[/tex]
[tex]16y^2=20x^2+12x+1[/tex]
Applying completing the square we get,
[tex]16y^2=20(x+\frac{3}{10})^2-\frac{4}{5}[/tex]
[tex]16y^2-20(x+\frac{3}{10})^2=-\frac{4}{5}[/tex]
[tex]\frac{16y^2}{-\frac{4}{5}}-{20(x+\frac{3}{10})^2}{-\frac{4}{5}}=1[/tex]
[tex]-\frac{y^2}{\frac{1}{4}}+{(x+\frac{3}{10})^2}{\frac{1}{25}}=1[/tex]
[tex]{(x+\frac{3}{10})^2}{\frac{1}{25}}-\frac{y^2}{\frac{1}{4}}=1[/tex]
This is in the form of hyperbola equation i.e. [tex]\frac{x^2}{a^2}-\frac{y^2}{b^2} =1[/tex]
Therefore, The given conic section is a hyperbola.
Hence, Option 2 is correct.
