Conic-Sections Question 547
Question: Equation of the latus rectum of the hyperbola $ {{(10x-5)}^{2}}+{{(10y-2)}^{2}}=9{{(3x+4y-7)}^{2}} $ is
Options:
A) $ y-\frac{1}{5}=-\frac{3}{4}( x-\frac{1}{2} ) $
B) $ x-\frac{1}{5}=-\frac{3}{4}( y-\frac{1}{2} ) $
C) $ y+\frac{1}{5}=-\frac{3}{4}( x+\frac{1}{2} ) $
D) $ x+\frac{1}{5}=-\frac{3}{4}( y+\frac{1}{2} ) $
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Answer:
Correct Answer: A
Solution:
[a] Given, hyperbola is $ {{(10x-5)}^{2}}+{{(10y-2)}^{2}}=9{{(3x+4y-7)}^{2}} $
$ \Rightarrow {{( x-\frac{1}{2} )}^{2}}+{{( y-\frac{1}{5} )}^{2}}=\frac{9}{4}{{( \frac{3x+4y-7}{5} )}^{2}} $
$ \Rightarrow $ Given curve is a hyperbola where focus is $ ( \frac{1}{2},\frac{1}{5} ) $ 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}( x-\frac{1}{2} ). $
BETA
