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Find the points on the curve `4x^2+9y^2=1`, where the tangents are perpendicular to the line `2y+x=0`. |
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Answer» The equation of the given line is `y = - (1)/(2)x` `:.` slope of this line `= - (1)/(2)`..(i) Let the required point be `(x_(1), y_(1))` Now, `4x^(2) + 9y^(2) = 1 rArr 8x + 18y .(dy)/(dx) = 0` `rArr (dy)/(dx) = ((-4x)/(9y)) rArr ((dy)/(dx))_((x_(1)"," y_(1))) = (-4x_(1))/(9 y_(1))` `:.` slope of the tangent at `(x_(1), y_(1)) = (-4x_(1))/(9y_(1))`...(ii) From (i) and (ii), we have `(-4x_(1))/(9y_(1)) xx (-(1)/(2)) = - 1 or y_(1) = - (2)/(9) x_(1)`..(iii) Since `(x_(1), y_(1))` lies on the curve `4x^(2) +9y^(2) = 1`, we have `4x_(1)^(2) + 9y_(1)^(2) = 1 rArr 4x_(1)^(2) + 9 xx (4)/(81) x_(1)^(2) = 1` [using (iii)] `:. x_(1)^(2) = (9)/(40) or x_(1) = +- (3)/(2sqrt10)` So, `y_(1) = +- (2)/(9) xx (3)/(2 sqrt10) = +- (1)/(3 sqrt10)` Hence, the required point are `((3)/(2 sqrt10), (1)/(3 sqrt10)) and ((-3)/(2 sqrt10) , (-1)/(3 sqrt10))` |
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