Find the coordinatesof the point on the curve xy=10 in the first quadrant such that the normal line passes through the orgin.
My work: Okay i believe this is an inverse variation. Therefore, y=10/x? Origin is @ (0,0) and i'm stuck to move on...
since y = 10/x, the slope of the tangent at (x,y) is -10/x^2
so the normal has slope x^2/10.
So, we have a line with slope y/x = (10/x)/(x^2/10) = 100/x^3
so, where does 10/x = 100/x^3?
x = √10
So, at (√10,√10), the slope is 1, and the normal line is just
y = x
Thanks I understand it now :D
To find the coordinates of the point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin, let's proceed step by step.
1. Start with the given curve equation, xy = 10.
2. We know that the equation of the normal line passing through a point (a, b) on the curve is given by the equation y - b = -1/m(x - a), where m is the slope of the tangent line to the curve at point (a, b).
3. To find the slope of the tangent line at point (a, b) on the curve, we differentiate the equation with respect to x. Differentiating xy = 10, we get d/dx(xy) = d/dx(10). Applying the product rule on the left side, we get y + xy' = 0. Solving for y', we have y' = -y/x.
4. Substituting the given curve equation xy = 10 into y' = -y/x, we get (-y/x) = -y/x. This equation implies that the slope of the tangent line at any point (x, y) on the curve is -1.
5. Now, let's find the coordinates of the point on the curve such that the normal line passes through the origin. Since the slope of the normal line passing through the origin is -1, the coordinates of the point on the curve will be of the form (a, 10/a), where a is a positive number.
6. Substitute y = 10/a in the equation xy = 10, we get xa = 10. Solving for a, we have a = 10/x. Since we are looking for points in the first quadrant, we need to find the value of x that is positive.
7. Now, since we know that a = 10/x, substituting it back into the point on the curve form (a, 10/a), we get (10/x, 10/(10/x)). Simplifying, we have (10/x, x).
Therefore, the point on the curve xy = 10 in the first quadrant, such that the normal line passes through the origin, has coordinates (10/x, x), where x is a positive number.
To find the coordinates of the point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin, we can follow these steps:
Step 1: Start with the equation of the curve, which is xy = 10. In the first quadrant, both x and y are positive.
Step 2: Assume a point on the curve with x-coordinate (x, y).
Step 3: We know that the normal line passes through the origin, which means it intersects the x-axis at (x, 0) and the y-axis at (0, y).
Step 4: The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the tangent line has slope dy/dx, the normal line will have a slope of -dx/dy.
Step 5: Solve the equation xy = 10 for y in terms of x: y = 10/x.
Step 6: Calculate the derivative of y with respect to x: dy/dx = -10/x^2.
Step 7: Determine the slope of the tangent line at the point (x, y) by substituting the x-coordinate into the derivative: dy/dx = -10/x^2 = -10/(x^2).
Step 8: Since the normal line is perpendicular to the tangent line, the slope of the normal line will be the negative reciprocal of the slope of the tangent line: -dx/dy = -(-x^2/10) = x^2/10.
Step 9: Set the slope of the normal line equal to the slope between (x, 0) and (0, y): x^2/10 = y/0. Rearranging the equation, we get: x^2/10 = 0.
Step 10: Solve for x: x^2 = 0. Taking the square root of both sides, we find x = 0.
Step 11: Since x = 0, substitute this value into the equation y = 10/x: y = 10/0 is undefined.
Therefore, there is no point on the curve xy = 10 in the first quadrant such that the normal line passes through the origin.