Question

The locus of the midpoint of the portion of a line of constant slope 'm' between two branches of the rectangular hyperbola xy=1 is

Answer 1

plz give nice explanations

Answer 2

Locus is y+mx = 0.

Step-by-step explanation:

Given:

Here, xy=1

Let equation of line be y = mx+c

Put in hyperbola x (mx+c)=1

⇒ $mx^2+cx-1=0$

Use Shridharacharya's formula

$x=\frac{-c\pm\sqrt{c^2+4m} }{2m}$

⇒ $x_1=\frac{-c+\sqrt{c^2+4m} }{2m}$ and $x_2=\frac{-c-\sqrt{c^2+4m} }{2m}$

Point lie on hyperbola , hence $y_1=\frac{1}{x_1}and y_2= \frac{1}{x_2}$

⇒ $y_1=\frac{2m }{-c+\sqrt{c^2+4m}} and y_2=\frac{2m }{-c-\sqrt{c^2+4m}}$

$(x_1,y_1)=(\frac{-c+\sqrt{c^2+4m} }{2m},\frac{2m }{-c+\sqrt{c^2+4m}})$

$(x_2,y_2)=(\frac{-c-\sqrt{c^2+4m} }{2m},\frac{2m }{-c-\sqrt{c^2+4m}})$

Let mid point be (h,k)

$h = (\frac{x_1+x_2}{2} )= \frac{\frac{-c+\sqrt{c^2+4m} }{2m}+\frac{-c-\sqrt{c^2+4m} }{2m}}{2}$

⇒ $h=-\frac{c}{2m}$             [1]

⇒ $k = (\frac{y_1+y_2}{2} = [tex]\frac{\frac{2m }{-c+\sqrt{c^2+4m}}+\frac{2m }{-c-\sqrt{c^2+4m}}}{2}$

= $\frac{2m}{2}[\frac{1}{\sqrt{c^2+4m}-c }-\frac{1}{\sqrt{c^2+4m}+c } ]$

= $m [\frac{\sqrt{c^2+4m}+c-\sqrt{c^2+4m}+c}{(\sqrt{c^2+4m}-c)(\sqrt{c^2+4m}+c }}]$

= $m[\frac{2c}{c^2+4m−c^2} ]$

= m[$\frac{2c}{4m}$]

⇒ $k = \frac{c}{2}$

c = 2k

Put in Eq (1)

$h=-\frac{c}{2m}$  

$h=-\frac{2k}{2m}$  

⇒ mh = −k

⇒ k + mh = 0

Hence locus is y+mx = 0