Question

The straight line \(\mathrm{y}=\mathrm{a}-\mathrm{x}\) touches the parabola \(\mathrm{x}^{2}=\mathrm{x}-\mathrm{y}\) if \(\mathrm{a}=\ldots \ldots \ldots .\) (a) \(-1\) (b) 0 (c) 1 (d) 2

Short answer

The value of 'a' for which the straight line touches the parabola is 1, which corresponds to answer choice (c).

Step-by-step solution

Find the point of intersection between the line and the parabola

To find the point of intersection, we can substitute the equation of the line y = a - x into the parabola equation x^2 = x - y: x^2 = x - (a - x) Now, we can solve for x and find the corresponding y coordinate:

Solve for x

From the previous step, we have the equation x^2 = x - (a - x). We can rearrange to create a quadratic equation: x^2 = 2x - a x^2 - 2x + a = 0

Find the corresponding y-coordinate

Recall the equation of the line y = a - x. Substituting x-coordinate we found in Step 2 into the line equation will give us the y-coordinate: y = a - x

Find the derivatives of line and parabola

Now that we have the coordinates of the intersection point, we can find the slopes of the line and parabola at the point where they touch. To do that, we will find the derivative of y with respect to x for both equations. For the line y = a - x, the derivative is constant: \( \frac{dy}{dx} = -1\) For the parabola x^2 = x - y, we need to find the derivative implicitly: \( \frac{d}{dx} (x^2) = \frac{d}{dx} (x - y) \) Applying the chain rule, we have: \( 2x = 1 - \frac{dy}{dx} \) Solving for \( \frac{dy}{dx} \), we get: \( \frac{dy}{dx} = 1 - 2x \)

Equate the slopes

To find the value of 'a' that makes the straight line touch the parabola, equate the slopes of the line and parabola: -1 = 1 - 2x Solving for x, we get: 2x = 2 x = 1 Now, substitute the x-value back into the line equation to find the corresponding y-value: y = a - x y = a - 1

Find the value of 'a'

Since the lines touch at (x, y) = (1, a - 1), we can substitute the point into the parabola equation x^2 = x - y: (1)^2 = 1 - (a - 1) Solving for 'a', we get: 1 = 2 - a a = 2 - 1 a = 1 Therefore, the value of 'a' for which the straight line touches the parabola is 1, which corresponds to answer choice (c).

Key concepts

Intersection of Line and Parabola

When a line intersects a parabola, it can do so at zero, one, or two points. In mathematical terms, these intersection points are solutions to the combined system of equations defined by the line and the parabola. By solving these equations simultaneously, we find the x and y coordinates where they meet. For our problem, we need to check if

• the line equation: \(y = a - x\)
• and the parabola equation: \(x^2 = x - y\)

are satisfied simultaneously by some values of \(x\) and \(y\).
We substitute \(y = a - x\) into \(x^2 = x - y\) to get a quadratic equation in terms of \(x\). This ultimately allows us to find the points of intersection.

Tangent Line

A tangent line is a special kind of line that just skims the surface of a curve at a single point. In our exercise, the line is only touching the parabola once, which means it's tangent. For any line to be tangent to a parabola, the slopes of the line and the parabola at the point of touch must be equal.To find where the tangent condition is met in our problem, we compute the derivatives of both the line and the parabola to compare their slopes. For the line, the slope is always \(-1\) because it's a linear equation structured as \(y = mx + c\). For the parabola, the derivative derived implicitly from \(x^2 = x - y\) yields the slope \(1 - 2x\). By equating these two slopes, we ensure that the line is indeed tangent to the parabola.

Derivative of Parabola

The derivative of a parabola helps find its instantaneous rate of change, which captures how steep the curve is at any point. In this problem, after implicitly differentiating the parabola equation, \(x^2 = x - y\), it gives the derivative:

• \(\frac{dy}{dx} = 1 - 2x\)

This equation tells us the slope of the parabola at any point along its curve. Particularly for the point where the parabola and line touch, this derivative becomes important. That's because, for the line to be tangent, its slope \(-1\) must match the parabola’s slope, \(1 - 2x\). The derivative also acts as a tool for finding critical values like local minima and maxima in broader calculus.

Quadratic Equation Solving

In solving the intersection of the line and the parabola, we end up with a quadratic equation. Quadratic equations come in the form of \(ax^2 + bx + c = 0\). Our problem leads to:

• \(x^2 - 2x + a = 0\)

This type of equation can usually be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), or by factoring if it's simple enough. In this exercise, though, the focus was on finding a specific value of \(a\) so that there was only one solution (indicating tangency). We achieved this by manipulating the equation and setting conditions for a double root, ensuring one solution.