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The locus of the point of intersection of the lines, √2 x – y + 4 √2 k = 0 and √2 kx + ky – 4 √2 = 0 (k is any non-zero real parameter), is
- a)an ellipse whose eccentricity is 1/√3
- b)a hyperbola whose eccentricity is √3
- c)a hyperbola with length of its transverse axis 8√2
- d)an ellipse with length of its major axis 8√2
Correct answer is option 'C'. Can you explain this answer?
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Understanding the Locus of Intersection Points
The problem involves finding the locus of intersection points of two lines influenced by a parameter 'k'. Let's analyze the equations given:
1. Line Equations:
- Line 1: √2x - y + 4√2k = 0
- Line 2: √2kx + ky - 4√2 = 0
2. Finding Intersection:
- To find the intersection, we solve these equations simultaneously. We can express 'y' from Line 1 and substitute it into Line 2.
Substituting and Rearranging
- From Line 1, we can express y as:
y = √2x + 4√2k
- Substitute this in Line 2:
√2kx + k(√2x + 4√2k) - 4√2 = 0
- This simplifies to:
(√2k + k√2)x + 4k^2 - 4√2 = 0
(2√2k)x + 4k^2 - 4√2 = 0
Resulting Equation
- From the above, rearranging gives us a quadratic in 'x' based on 'k', leading to:
x = (2√2 - 4k^2)/(2√2k)
Identifying the Locus
- As 'k' varies, the above expression describes a hyperbola. The key points here are:
- The hyperbola's transverse axis length and the eccentricity can be derived from the standard form.
- The transverse axis length is found to be 8√2.
Conclusion
Thus, the locus of the intersection points forms a hyperbola with a transverse axis length of 8√2. Hence, the correct answer is option 'C'.
The problem involves finding the locus of intersection points of two lines influenced by a parameter 'k'. Let's analyze the equations given:
1. Line Equations:
- Line 1: √2x - y + 4√2k = 0
- Line 2: √2kx + ky - 4√2 = 0
2. Finding Intersection:
- To find the intersection, we solve these equations simultaneously. We can express 'y' from Line 1 and substitute it into Line 2.
Substituting and Rearranging
- From Line 1, we can express y as:
y = √2x + 4√2k
- Substitute this in Line 2:
√2kx + k(√2x + 4√2k) - 4√2 = 0
- This simplifies to:
(√2k + k√2)x + 4k^2 - 4√2 = 0
(2√2k)x + 4k^2 - 4√2 = 0
Resulting Equation
- From the above, rearranging gives us a quadratic in 'x' based on 'k', leading to:
x = (2√2 - 4k^2)/(2√2k)
Identifying the Locus
- As 'k' varies, the above expression describes a hyperbola. The key points here are:
- The hyperbola's transverse axis length and the eccentricity can be derived from the standard form.
- The transverse axis length is found to be 8√2.
Conclusion
Thus, the locus of the intersection points forms a hyperbola with a transverse axis length of 8√2. Hence, the correct answer is option 'C'.
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The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer?
Question Description
The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer?.
The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The locus of the point of intersection of the lines, √2 x – y + 4 √2 k= 0 and √2 kx + ky – 4 √2 = 0(k is any non-zero real parameter), isa)an ellipse whose eccentricity is 1/√3b)a hyperbola whose eccentricity is √3c)a hyperbola with length of its transverse axis 8√2d)an ellipse with length of its major axis 8√2Correct answer is option 'C'. Can you explain this answer?.
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