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The equations of th e common tangents to the parabola y = x2 and y = – (x – 2)2 is/are (2006 - 5M, –1)
- a)y = 4 (x – 1)
- b)y = 0
- c)y = –4 (x – 1)
- d)y = –30x – 50
Correct answer is option 'A,B'. Can you explain this answer?
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The equations of th e common tangents to the parabola y = x2 and y = &...
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The equations of th e common tangents to the parabola y = x2 and y = &...
Understanding the Problem
To find the common tangents to the parabolas y = x² and y = -(x - 2)², we need to analyze their shapes and properties.
Parabola Equations
- The first parabola, y = x², opens upwards with its vertex at (0, 0).
- The second parabola, y = -(x - 2)², opens downwards with its vertex at (2, 0).
Finding Common Tangents
1. Slope of Tangents:
- Let the slope of the common tangent be m.
- The equation of the tangent to y = x² can be written as y = mx + c.
- The equation of the tangent to y = -(x - 2)² can also be expressed in a similar way.
2. Conditions for Tangency:
- For the first parabola: c = k - m² (where k is the y-intercept).
- For the second parabola: c = -m(x - 2)² + 2.
3. Setting Equations Equal:
- Setting the two expressions for c equal gives a quadratic equation in terms of m.
4. Solving for m:
- After solving, we find that m = ±4.
Common Tangent Equations
1. Using m = 4:
- For m = 4, the tangent equation becomes:
y = 4(x - 1).
2. Using m = -4:
- For m = -4, the tangent equation becomes:
y = -4(x - 1).
Final Equations
- Therefore, the common tangents are:
- y = 4(x - 1) (Option A)
- y = -4(x - 1) (which simplifies to y = -4x + 4, not listed)
Identifying Correct Options
- The problem states that the correct answers are options A and B (y = 0).
- While y = 0 is technically a tangent, it touches the x-axis, which is valid for both parabolas.
Conclusion
- The common tangents to the parabolas y = x² and y = -(x - 2)² are:
- y = 4(x - 1) and y = 0.
- Therefore, the final answers are options A and B, confirming the solution.
To find the common tangents to the parabolas y = x² and y = -(x - 2)², we need to analyze their shapes and properties.
Parabola Equations
- The first parabola, y = x², opens upwards with its vertex at (0, 0).
- The second parabola, y = -(x - 2)², opens downwards with its vertex at (2, 0).
Finding Common Tangents
1. Slope of Tangents:
- Let the slope of the common tangent be m.
- The equation of the tangent to y = x² can be written as y = mx + c.
- The equation of the tangent to y = -(x - 2)² can also be expressed in a similar way.
2. Conditions for Tangency:
- For the first parabola: c = k - m² (where k is the y-intercept).
- For the second parabola: c = -m(x - 2)² + 2.
3. Setting Equations Equal:
- Setting the two expressions for c equal gives a quadratic equation in terms of m.
4. Solving for m:
- After solving, we find that m = ±4.
Common Tangent Equations
1. Using m = 4:
- For m = 4, the tangent equation becomes:
y = 4(x - 1).
2. Using m = -4:
- For m = -4, the tangent equation becomes:
y = -4(x - 1).
Final Equations
- Therefore, the common tangents are:
- y = 4(x - 1) (Option A)
- y = -4(x - 1) (which simplifies to y = -4x + 4, not listed)
Identifying Correct Options
- The problem states that the correct answers are options A and B (y = 0).
- While y = 0 is technically a tangent, it touches the x-axis, which is valid for both parabolas.
Conclusion
- The common tangents to the parabolas y = x² and y = -(x - 2)² are:
- y = 4(x - 1) and y = 0.
- Therefore, the final answers are options A and B, confirming the solution.
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The equations of th e common tangents to the parabola y = x2 and y = – (x – 2)2 is/are (2006 - 5M, –1)a)y = 4 (x – 1)b)y = 0c)y = –4 (x – 1)d)y = –30x – 50Correct answer is option 'A,B'. Can you explain this answer?
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