The graph of the equation x2 − xy + y2 = 9 is an ellipse. Find the lines tangent to this curve at the two points where it intersects the x-axis. Show that these lines are parallel.
x^2 - xy + y^2 = 9
2x - x dy/dx - y + 2y dy/dx = 0
dy/dx(2y - x) = y - 2x
dy/dx = (y-2x)/(2y-x)
at x-intercept: y = 0
x^2 - 0 + 0 = 9
x = ± 3
when x = 3, dy/dx = (0-6)/(0-3) = 2
equation of tangent:
y = 2(x-3)
y = 2x - 6
when x = -3 , dy/dx = (0 + 6)/(0+3) = 2
equation at (-3,0)
y = 2(x+3)
y = 2x + 6
since both lines have a slope of 2 , they are parallel
To find the lines tangent to the curve at the points where it intersects the x-axis, we first need to find the coordinates of these points.
To do this, we set y = 0 in the equation x^2 − xy + y^2 = 9 and solve for x:
x^2 − x(0) + (0)^2 = 9
x^2 = 9
Taking the square root of both sides, we get:
x = ±3
So, the curve intersects the x-axis at the points (3, 0) and (-3, 0).
Now, let's find the slope of the curve at each of these points. We differentiate the equation x^2 − xy + y^2 = 9 implicitly with respect to x:
2x - y - x(dy/dx) + 2y(dy/dx) = 0
Rearranging this equation, we get:
(dy/dx) = (2x - y) / (x - 2y)
Substituting the coordinates of the points (3, 0) and (-3, 0), we can find the slopes at these points:
At x = 3:
(dy/dx) = (2(3) - 0) / (3 - 2(0)) = 6/3 = 2
At x = -3:
(dy/dx) = (2(-3) - 0) / (-3 - 2(0)) = -6/-3 = 2
We can see that the slopes at both points are equal and hence, the lines tangent to the curve at these points are parallel.
Therefore, the lines tangent to the curve x^2 − xy + y^2 = 9 at the two points where it intersects the x-axis are parallel.
To find the lines that are tangent to the curve at the points where it intersects the x-axis, we first need to find the points of intersection.
When the curve intersects the x-axis, the y-coordinate of those points is 0. Thus, we can substitute y = 0 into the equation x^2 - xy + y^2 = 9 to solve for x.
x^2 - x(0) + (0)^2 = 9
x^2 = 9
x = ±√9
x = ±3
So, the curve intersects the x-axis at the points (-3, 0) and (3, 0).
Next, we need to find the slopes at these points. To do this, we take the derivative of the given equation implicitly with respect to x.
Differentiating x^2 - xy + y^2 = 9 with respect to x, we get:
2x - y - xy' + 2yy' = 0
Since we are interested in the slopes at the points of intersection with the x-axis, we substitute y = 0 in the derivative equation:
2x - (0) - x(0) + 2(0)y' = 0
2x = 0
x = 0
At x = 0, the slope is given by y' = -2x/(2y).
Substituting x = 0 and y = 0 into y', we find that the slope is undefined.
This means that the curve has vertical tangents at the points of intersection with the x-axis. The equations of vertical lines are of the form x = c, where c is a constant.
Hence, the lines tangent to the curve at the points (-3, 0) and (3, 0) are x = -3 and x = 3, respectively.
Since both lines have equations of the form x = c, where c is a constant, their slopes are undefined. Therefore, the lines are parallel.