Find the coordinates of the point on the curve xy=6 in quadrant I where the normal to the curve passes through the origin.
To find the coordinates of the point on the curve xy = 6 in quadrant I where the normal to the curve passes through the origin, we can start by finding the equation of the normal at any point on the curve.
Step 1: Differentiate the equation xy = 6 implicitly with respect to x:
d(xy)/dx = d(6)/dx
y + x * dy/dx = 0
Step 2: Rearrange the equation to solve for dy/dx:
dy/dx = -y / x
Step 3: At the point (a, b) on the curve, the slope of the tangent line is dy/dx = -b / a. Since the normal is perpendicular to the tangent line, the slope of the normal is the negative reciprocal of the tangent line's slope, given by -a / b.
Step 4: Given that the normal passes through the origin (0, 0), we can write the equation of the normal as:
(y - 0) = (-a / b) * (x - 0)
y = (-a / b) * x
Step 5: Now, substituting the value of xy = 6 in the equation of the normal, we have:
(-a / b) * x = 6 / x
Step 6: Simplify the equation by multiplying both sides by x and b:
-a * x = 6 * b
Step 7: Solve for x:
x = -6 * b / a
Step 8: Substitute the value of x into the equation of the curve xy = 6:
(-6 * b / a) * y = 6
y = -a / b
Step 9: Substitute the value of y into the equation of the normal to find the x-coordinate of the point:
(-a / b) * x = 6 / x
(a^2 / b) = 6
a^2 = 6b
Step 10: Now, substitute the value of a^2 from step 9 into the equation from step 7 to solve for x:
x = -6 * b / a
x = -6 * b / sqrt(6b)
x = -6 * sqrt(b)
Therefore, the coordinates of the point on the curve xy = 6 in quadrant I where the normal to the curve passes through the origin are (-6 * sqrt(b), -a / b).
To find the coordinates of the point on the curve xy = 6 in quadrant I where the normal to the curve passes through the origin, we need to follow these steps:
Step 1: Understand the Problem
The equation xy = 6 represents a curve, and we are looking for a specific point on this curve. The normal to the curve is a line perpendicular to the tangent of the curve at a given point. We need to find the point on the curve where the normal passes through the origin.
Step 2: Define Variables
Let's assume that the point on the curve is (a, b). We need to find the values of a and b.
Step 3: Find the Derivative
To find the tangent to the curve at point (a, b), we need to differentiate the equation xy = 6 with respect to x. Taking the derivative implicitly, we get:
d(xy)/dx = d(6)/dx
Using the product rule of differentiation, we can rewrite this as:
y + x * dy/dx = 0
Step 4: Find the Slope of the Curve
Now, let's find the slope of the curve at the point (a, b). To do this, we'll substitute x = a and y = b into the derivative equation we obtained in step 3:
b + a * dy/dx = 0
This equation gives us the slope of the curve at point (a, b).
Step 5: Find the Slope of the Normal
The slope of the normal to the curve is the negative reciprocal of the slope of the curve at the given point. Let's call this slope m_n. So, we have:
m_n = -1 / (dy/dx)
Step 6: Find the Equation of the Normal
We now have the slope of the normal (m_n) and the point it passes through (the origin, which is (0, 0)). Using the point-slope form of a linear equation, we can write the equation of the normal line:
y - 0 = m_n * (x - 0)
Simplifying, we get:
y = m_n * x
Step 7: Solve for x
Now, we substitute the equation of the curve xy = 6 into the equation of the normal line. This gives us:
b = m_n * a
Since we are working in the first quadrant, both a and b are positive. Hence, we can replace the product b = m_n * a, because we know b > 0 and m_n < 0.
Step 8: Substitute Equation
We substitute m_n = -1 / (dy/dx) from step 5 into the equation from step 7 to find:
b = (-1 / (dy/dx)) * a
Step 9: Solve for a
We substitute the equation xy = 6 into the equation derived from step 8:
(-1 / (dy/dx)) * a = 6 / a
Simplifying, we get:
-a^2 = 6 / (dy/dx)
Step 10: Find dy/dx
We return to the derivative equation from step 3:
y + x * dy/dx = 0
We can rearrange this equation to solve for dy/dx:
dy/dx = -y / x
Step 11: Substitute into the Equation
We substitute the expression for dy/dx (-y / x) into the equation from step 10:
-a^2 = 6 / (-y / x)
Simplifying further:
-a^2 = -6x / y
Step 12: Find y in Terms of a
From the equation xy = 6, we express y in terms of a:
y = 6 / a
Substituting this into the equation from step 12, we get:
-a^2 = -6x / (6 / a)
Simplifying:
-a^2 = -ax
Dividing by -a:
a = x
Step 13: Solve for a
Finally, we substitute a = x into the equation xy = 6:
x * y = 6
Since a = x, we can rewrite this as:
a * y = 6
Substituting y = 6 / a:
a * (6 / a) = 6
Simplifying, we find:
6 = 6
This equation holds true for all positive values of a. So, there is no unique solution for the point (a, b) on the curve where the normal passes through the origin. The curve xy = 6 does not possess such a point in quadrant I.