Question
Equation of the latus rectum of the hyperbola $${\left( {10x - 5} \right)^2} + {\left( {10y - 2} \right)^2} = 9{\left( {3x + 4y - 7} \right)^2}{\text{ is :}}$$
A.
$$y - \frac{1}{5} = - \frac{3}{4}\left( {x - \frac{1}{2}} \right)$$
B.
$$x - \frac{1}{5} = - \frac{3}{4}\left( {y - \frac{1}{2}} \right)$$
C.
$$y + \frac{1}{5} = - \frac{3}{4}\left( {x + \frac{1}{2}} \right)$$
D.
$$x + \frac{1}{5} = - \frac{3}{4}\left( {y + \frac{1}{2}} \right)$$
Answer :
$$y - \frac{1}{5} = - \frac{3}{4}\left( {x - \frac{1}{2}} \right)$$
Solution :
Given, hyperbola is
$$\eqalign{
& {\left( {10x - 5} \right)^2} + {\left( {10y - 2} \right)^2} = 9{\left( {3x + 4y - 7} \right)^2} \cr
& \Rightarrow {\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{5}} \right)^2} = \frac{9}{4}{\left( {\frac{{3x + 4y - 7}}{5}} \right)^2} \cr} $$
$$ \Rightarrow $$ Given curve is a hyperbola where focus is $$\left( {\frac{1}{2},\,\frac{1}{5}} \right)$$ and directrix is $$3x + 4y - 7 = 0.$$ Latus rectum is a line passing through the focus and parallel to the directrix.
$$ \Rightarrow $$ Equation of the latus rectum is $$y - \frac{1}{5} = - \frac{3}{4}\left( {x - \frac{1}{2}} \right).$$