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The equation of the line passing through the centre of a rectangular hyperbola is `x-y-1=0`. If one of its asymptotoes is `3x-4y-6=0`, the equation of the other asymptote isA. `4x+3y+17=0`B. `4x-3y+8=0`C. `3x-2y+15=0`D. none of these |
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Answer» The point of intersection of the line `x-y-1=0`, which passes through the centre of the hyperbola, and the asymptote `3x-4y-6=0` is the centre of the hyperbola. So, its coordinates are `(-2,-3)`. Since asymptotes of a rectangular hyperbola are always at right angle. So, required asymptotes is perpendicular to the given asymptote and passes through the centre `(-2,-3)` of the hypebola and hence its equation is `y+3=-(4)/(3)(x+2)` or, `4x+3y+17=0` |
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