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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

If the tangents are drawn to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=12\) at the point where it meets the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-5 \mathrm{x}+3 \mathrm{y}-2=0\), then the point of intersection of these tangent is (a) \((6,-6)\) (b) \([6,(18 / 5)]\) (c) \([-6,(18 / 5)]\) (d) \([6,(18 / 5)]\)

Let \(\mathrm{C}\) be the centre of the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-4 \mathrm{y}-20=0\). If the tangents at the point \(\mathrm{A}(1,7)\) and \(\mathrm{B}(4,-2)\) on the circle meet at point \(\mathrm{D}\). Then area of the quadrilateral \(\mathrm{ABCD}\) is \(\ldots \ldots\) (a) 150 sq. units (b) 100 sq. units (c) 75 sq. units (d) 50 sq. units

A square is inscribed in the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+4 \mathrm{y}-93=0\) Its sides are parallel to the coordinate axes. Then one vertex of the square is (a) \((1+\sqrt{2},-2)\) (b) \((1-\sqrt{2},-2)\) (c) \((1,-2+\sqrt{2})\) (d) none of these

The locus of a point \(\mathrm{P}(\alpha, \beta)\) moving under the condition that the line \(\mathrm{y}=\alpha \mathrm{x}+\beta\) is a tangent to the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) is (a) a circle (b) a parabola (c) an ellipse (d) a hyperbola

The number of common tangents to the circles \(\mathrm{x}^{2}+\mathrm{y}^{2}=4\) and \(x^{2}+y^{2}-6 x-8 y-24=0\) is \(\ldots \ldots\) (a) 0 (b) 1 (c).2 (d) None of these
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