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The equation of tangent to hyperbola `4x^2-5y^2=20` which is parallel to `x-y=2` is (a) `x-y+3=0` (b) `x-y+1=0` (c) `x-y=0` (d) `x-y-3=0`A. `x-y-3=0`B. `x-y+9=0`C. `x-y+1=0`D. `x-y+7=0` |
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Answer» Given equation of hyperbola is `4x^(2)-5y^(2)=20` which can be rewritten as ` rArr (x^(2))/(5)-(y^(2))/(4)=1` The line `x-y=2` has slope , `m=1` `therefore` Slope of tangent parallel to this line = 1 We know equation of tangent to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` having slope m is given by `y=mx pm sqrt(a^(2)m^(2)-b^(2))` Here, `a^(2)=5, b^(2)=4 and m=1` ` therefore` Required equation of tangent is `rArr y=xpm sqrt(5-4)` `rArr y=x pm 1 rArr x-ypm 1=0` |
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