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

\[x-\frac{1}{5}=-\frac{3}{4}\left( y-\frac{1}{2} \right)\]

\[y-\frac{1}{5}=-\frac{3}{4}\left( x-\frac{1}{2} \right)\]

\[y+\frac{1}{5}=-\frac{3}{4}\left( x+\frac{1}{2} \right)\]

\[x+\frac{1}{5}=-\frac{3}{4}\left( y+\frac{1}{2} \right)\]

Equation of the latus rectum of the hyperbola\[{{(10x-5)}^{2}}+{{(10y-

1.	Equation of the latus rectum of the hyperbola\[{{(10x-5)}^{2}}+{{(10y-2)}^{2}}=9{{(3x+4y-7)}^{2}}\] 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» B. \[x-\frac{1}{5}=-\frac{3}{4}\left( y-\frac{1}{2} \right)\]

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

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