Find the points on the curve y = 3x 2 -9x +8 at which the tangents are equally inclined with the axes . ans : (5/3 , 4/3) and (4/3 , 4/3) how come two answers of this qs.
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y = 3 x² - 9 x + 8
dy/dx = m = slope of tangent at (x,y) = 6x - 9
When the tangent line intersects axes x and y, the angles it makes with both axes are same. In the triangle formed with origin and axes, sum of these two internal angles is 90 deg. So internal angle is 45 deg.
If the tangent is in quadrant 1 or 3, then slope = -1 ie., angle is 135 deg. If the tangent is in quadrant 2 or 4, then slope = 1 ie, angle is 45 deg.
So the points:
m = 1, 6x - 9 = 1 => x = 5/3 => y = 3 (5/3)² - 9 *5/3 +8 = 4/3
m = -1, 6x - 9 = -1 => x = 4/3 => y = 3 (4/3)² - 9*4/3 +8 = 4/3
so we have two points.
dy/dx = m = slope of tangent at (x,y) = 6x - 9
When the tangent line intersects axes x and y, the angles it makes with both axes are same. In the triangle formed with origin and axes, sum of these two internal angles is 90 deg. So internal angle is 45 deg.
If the tangent is in quadrant 1 or 3, then slope = -1 ie., angle is 135 deg. If the tangent is in quadrant 2 or 4, then slope = 1 ie, angle is 45 deg.
So the points:
m = 1, 6x - 9 = 1 => x = 5/3 => y = 3 (5/3)² - 9 *5/3 +8 = 4/3
m = -1, 6x - 9 = -1 => x = 4/3 => y = 3 (4/3)² - 9*4/3 +8 = 4/3
so we have two points.
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