Question

The length of the latus rectum of the conic $\frac{5}{r}=2+3\cos \theta +4\sin \theta$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Convert into Cartesian coordinates by using $x=r\cos \theta ,y=r\sin \theta ,r=\sqrt{x^2+y^2}$

$5=2r+3r\cos \theta +4r\sin \theta$

$\Rightarrow 5=2\sqrt{x^2+y^2}+3x+4y$

$\Rightarrow 3x+4y-5=-2\sqrt{x^2+y^2}$

$\Rightarrow x^2+y^2=\frac{25}{4}{\left(\frac{3x+4y-5}{5}\right)}^2$

Hence, it's a conic with eccentricity$=\frac{5}{2}$, focus$(0,0)$ and directrix $3x+4y-5=0$

Focal parameter = perpendicular distance between focus and directrix = $\left|\frac{-5}{5}\right|=1$

Latus rectum $=2\times$ focal parameter $\times$ eccentricity

$=2\times 1\times \frac{5}{2}=5$