Question

Angle between the parabolas $y^2=4b(x-2a+b)$ and $x^2+4a(y-2b-a)=0$ at the common end of their latus rectum, is?

Answer 1

$y^2=4b(x-2a+b),x^2+4a(y-2b-a)=0$

this is in the form

$y^2=4ax$

$x-2a+b\le b$

$x=2a$

End of latus rectum $(2a,2b)(2a,-2b)$

Coordinates Slope

$2y{y}^{\prime }=4b$

$y^{\prime }=\frac{2b}{y}$

$y^{\prime }=\frac{2b}{2b}=1$

End of latus rectum

$y-2b-a=-a$

$y=2b$

$(-2a,2b)(2a,2b)$

Slope $2x+4a{y}^{\prime }=0$

$y^{\prime }=\frac{-x}{2a}$

$y^{\prime }=\frac{-2a}{2a}=-1$

Product of slopes $=|x-|=-1$

$\therefore$ They are perpendicular

So their angle is $\frac{\pi }{2}$ (end of latus rectum)

Let us check option

$A\left){\tan }^{-1}\right(1)=\frac{\pi }{4}$

$C\left){\tan }^{-1}\right(\sqrt{3})=\frac{\pi }{3}$

$D\left){\tan }^{-1}\right(2)=+{\tan }^{-1}(3)$

${\tan }^{-1}\frac{2+3}{1-2.3}$

${\tan }^{-1}\frac{5}{-5}=\frac{3\pi }{4}$

So our correct option is $B$ i.e;

${\tan }^{-1}+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$

${\tan }^{-1}1+{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)$

${\tan }^{-1}1+{\tan }^{-1}\frac{5}{5}$

${\tan }^{-1}1+{\tan }^{-1}1$

$\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$