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Chapter 15: Problem 1387

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

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

Step by step solution

01

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:
02

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
03

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
04

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: \( \frac{dy}{dx} = 1 - 2x \)
05

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
06

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).

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Most popular questions from this chapter

Let \(E\) be the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1\) and \(C\) be the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\). Let \(\mathrm{P}\) and \(\mathrm{Q}\) be the point \((1,2)\) and \((2,1)\) respe. Then (a) P lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\) (b) P lies inside both \(\mathrm{C}\) and \(\mathrm{E}\) (c) \(Q\) lies outside both \(\mathrm{C}\) and \(\mathrm{E}\) (d) Q lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\)

The lines \(2 \mathrm{x}-3 \mathrm{y}-5=0\) and \(3 \mathrm{x}-4 \mathrm{y}-7=0\) are diameters of a circle of area 154 square units then the equation of the circle is (a) \(x^{2}+y^{2}+2 x-2 y-62=0\) (b) \(x^{2}+y^{2}+2 x-2 y-47=0\) (c) \(x^{2}+y^{2}-2 x+2 y-47=0\) (d) \(x^{2}+y^{2}-2 x+2 y-62=0\)

The eccentricity of an ellipse, with its centre at the origin, is \((1 / 2)\). If one of the directories is \(\mathrm{x}=4\), then equation of an ellipse is (a) \(3 \mathrm{x}^{2}+4 \mathrm{y}^{2}=1\) (b) \(3 x^{2}+4 y^{2}=12\) (c) \(4 x^{2}+3 y^{2}=12\) (d) \(4 x^{2}+3 y^{2}=1\)

If \(\mathrm{x}+\mathrm{y}+1=0\) touches the parabola \(\mathrm{y}^{2}=\mathrm{ax}\) then \(\mathrm{a}=\ldots \ldots \ldots\) (a) 8 (b) 6 (c) 4 (d) 2

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