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Find the equation of the tangents to the curve `2x^(2) + 3y^(2) = 14`, parallel to the line `x + 3y = 4` |
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Answer» Correct Answer - `x + 3y = 7, x + 3y = -7` The given lies is `x + 3y = 4 rArr y = - (1)/(3) x + (4)/(3)` Slope of tangent = slope of given lines `= (-1)/(3)` Let the point of contact be `(x_(1), y_(1))` Now, `2x^(2) + 3y^(2) = 14 rArr 4x + 6y (dy)/(dx) = 0 rArr (dy)/(dx) = (-4x)/(6y) = (-2x)/(3y) rArr ((dy)/(dx))_((x_(1)"," y_(1)) = (-2x)/(3y_(1))` `:. (-2x_(1))/(3y_(1)) = (-1)/(3) rArr 2x_(1) = y_(1)` Also, `(x_(1), y_(1))` lies on the given curve `:. 2x_(1)^(2) + 3y_(1)^(2) = 14 rArr 2x_(1)^(2) + 3 xx 4x_(1)^(2) = 14 rArr x_(1)^(2) = 1 rArr x_(1) = +- 1` `(x_(1) = 1 rArr y_(1) = 2) and (x_(1) = -1 rArr y_(1) = -2)` So, the points of contact are (1, 2) and `(-1, -2)` The respective tangents are `(y -2)/(x -1) = (-1)/(3) adn (y +2)/(x +1) = (-1)/(3)` `rArr 3y - 6 = -x + 1 and 3y + 6 = -x - 1 rArr 3y + x = 7 and 3y + x = -7` |
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