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) \[x+\frac{1}{5}=-\frac{3}{4}\left( y+\frac{1}{2} \right)\]
Correct Answer: A
Solution :
[a] Given, hyperbola is \[{{(10x-5)}^{2}}+{{(10y-2)}^{2}}=9{{(3x+4y-7)}^{2}}\] \[\Rightarrow {{\left( x-\frac{1}{2} \right)}^{2}}+{{\left( y-\frac{1}{5} \right)}^{2}}=\frac{9}{4}{{\left( \frac{3x+4y-7}{5} \right)}^{2}}\] \[\Rightarrow \] 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. \[\Rightarrow \]Eq. of the latus rectum is \[y-\frac{1}{5}=-\frac{3}{4}\left( x-\frac{1}{2} \right).\]
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