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Chapter 15: Problem 1399

If the line \(\mathrm{y}=1-\mathrm{x}\) touches the curve \(\mathrm{y}^{2}-\mathrm{y}+\mathrm{x}=0\), then the point of contact is (a) \((0,1)\) (b) \((1,0)\) (c) \((1,1)\) (d) \([(1 / 2),(1 / 2)]\)

Short Answer

Expert verified
The correct point of contact between the line \(y = 1 - x\) and the curve \(y^{2} - y + x = 0\) is (b) \((1, 0)\).

Step by step solution

01

Understanding the Line Equation

The given line equation is y = 1 - x. This is a straight line which has a slope of -1 and passes through the point (0,1).
02

Understanding the Curve Equation

The given curve equation is y² - y + x = 0. It is a quadratic equation in y with respect to x.
03

Substituting Line Equation into Curve Equation

We need to find the point where the two equations intersect. Therefore, we will substitute the equation of the line y = 1 - x, into the curve equation: \[(1-x)^{2} - (1-x) + x = 0\]
04

Solve the Quadratic Equation for x

Now, let's expand and simplify the quadratic equation to solve for x: \[1-2x + x^{2} - 1 + x + x = 0\] \[ x^{2} - x = 0\] Now, let's solve for the x by factoring out x: \[x( x - 1 ) = 0\] This gives us two possible values of x: 1. x = 0 2. x = 1
05

Solve for y using the Line Equation

Next, let's find the corresponding values for y using the line equation y = 1 - x: 1. For x = 0, y = 1 - 0 = 1 2. For x = 1, y = 1 - 1 = 0 Thus, we have two possible points of contact (0,1) and (1,0).
06

Determine which Point is the Correct Point of Contact

We know that the tangent should touch the curve at exactly one point. So, let's analyze the two points of contact we found: 1. For the point of contact (0,1): Slope of the line is -1, and the curve y² - y + x = 0 becomes y² - y = 0 (since x=0), which is y(y - 1) = 0. After analyzing the curve, we can see that the curve touches the x-axis at x=(1/2). So, the slope at this point is not -1. 2. For the point of contact (1,0): The curve y² - y + x = 0 becomes (y - 0)² + 1 = 0 (since y=0), which is y² + 1 = 0. After analyzing the curve, we can see that the line y=1-x is tangent to the curve at (1,0). Hence, the correct point of contact is (1,0), which means the answer is (b) (1,0).

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