Question

Equation of one of the latus rectum of the hyperbola $(10x-5{)}^2+(10y-2{)}^2=9(3x+4y-7{)}^2$ is

• A. $30x+40y-23=0$
• B. $40x+30y-23=0$
• C. $30x+40y+23=0$
• D. $40x+30y+23=0$

Answer 1

The correct option is A $30x+40y-23=0$

Given, hyperbola is
$(10x-5{)}^2+(10y-2{)}^2=9(3x+4y-7{)}^2$

$\Rightarrow 100[{(x-\frac{1}{2})}^2+{(y-\frac{1}{5})}^2]=9(3x+4y-7{)}^2$

$\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{1}{5})}^2=\frac{9}{4}{\left(\frac{3x+4y-7}{\sqrt{2}5}\right)}^2$

Which represents hyperbola having one focus $S(\frac{1}{2},\frac{1}{5})$ and directrix equation as $3x+4y-7=0$
Let equation of latus rectum be $3x+4y+\lambda =0$
As it passes through $(\frac{1}{2},\frac{1}{5})$
$\Rightarrow \lambda =\frac{-23}{10}$

Hence, equation of the latus rectum is $30x+40y-23=0$