AP EAMCET · Maths · Hyperbola
If the equation of one asymptote of the hyperbola \(14 x^2+38 x y+20 y^2+x-7 y-91=0\) is \(7 x+5 y-3=0\), then the other asymptote is
- A \(2 x-4 y+1=0\)
- B \(2 x+4 y+1=0\)
- C \(2 x-4 y-1=0\)
- D \(2 x+4 y-1=0\)
Answer & Solution
Correct Answer
(B) \(2 x+4 y+1=0\)
Step-by-step Solution
Detailed explanation
Given hyperbola, On factorising \(14 x^2+38 x y+20 y^2\), we get \[ =(7 x+5 y)(2 x+4 y) \] One of the asymptote is \(7 x+5 y-3=0\) Then, let other asymptote is \(2 x+4 y+k=0\) So, on combining On equating the coefficient of \(x\) from Eqs. (i) and (ii), we get…
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