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Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 is
- a)| z - (2 + 3i) | = 1
- b)| z - (2 + 3i) | = 4
- c)| z - (1 + i) | = 2
- d)| z - (1 + i) | = 1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4...
represents a hyperbola with foci (1, 2) and (3, 4) and length of transverse axis = 2.
∴ 2a = 2 ⇒ a = 1
∵ Feet of perpendiculars from foci on any tangent lie on auxilliary circle of the hyperbola.
∴ Locus will be auxilliary circle.
∴ Centre = mid point of foci = (2, 3)
and radius = semi transverse axis = 1
∴ Equation of auxilliary circle is |z - (2 + 3i) |= 1
Most Upvoted Answer
Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4...
To find the locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic, we can start by understanding the given conic equation and its properties.
Given equation: | z - (1 + 2i) | - | z - (3 + 4i) | = 2
This equation represents a hyperbola with foci at (1, 2) and (3, 4) and the difference of distances from any point on the hyperbola to the two foci is constant and equal to 2.
To find the locus of the feet of perpendiculars, let's consider a point P on the hyperbola with coordinates (x, y).
Now, let's find the equation of the tangent at point P. The equation of a tangent to a conic can be written in the form: zz̅ - α(z + z̅) + c = 0, where α is a complex constant and c is a real constant.
Let's substitute the coordinates of point P in the equation of the hyperbola to find α.
| (x - 1) + i(y - 2) | - | (x - 3) + i(y - 4) | = 2
Simplifying this equation, we get:
√((x - 1)² + (y - 2)²) - √((x - 3)² + (y - 4)²) = 2
Squaring both sides of the equation and simplifying, we get:
(x - 1)² + (y - 2)² - 2√((x - 1)² + (y - 2)²)√((x - 3)² + (y - 4)²) + (x - 3)² + (y - 4)² = 4
Expanding and combining like terms, we get:
2x² + 2y² - 8x - 12y + 14 = 2√((x - 1)² + (y - 2)²)√((x - 3)² + (y - 4)²)
Now, we can compare this equation with the general equation of a tangent to a conic:
zz̅ - α(z + z̅) + c = 0
By comparing coefficients, we can equate the real and imaginary parts separately to find α.
From the real part: 2x² + 2y² - 8x - 12y + 14 = 0
From the imaginary part: -2{x(x - 4) + y(y - 6)} = 0
Simplifying both equations, we get:
x² - 4x + y² - 6y + 2 = 0
This is the equation of the tangent at any point (x, y) on the hyperbola.
Now, let's find the coordinates of the feet of perpendiculars from points (1, 2) and (3, 4) on this tangent.
The feet of perpendiculars can be found by substituting the coordinates of the points into the equation of the tangent and solving for x and y.
For point (
Given equation: | z - (1 + 2i) | - | z - (3 + 4i) | = 2
This equation represents a hyperbola with foci at (1, 2) and (3, 4) and the difference of distances from any point on the hyperbola to the two foci is constant and equal to 2.
To find the locus of the feet of perpendiculars, let's consider a point P on the hyperbola with coordinates (x, y).
Now, let's find the equation of the tangent at point P. The equation of a tangent to a conic can be written in the form: zz̅ - α(z + z̅) + c = 0, where α is a complex constant and c is a real constant.
Let's substitute the coordinates of point P in the equation of the hyperbola to find α.
| (x - 1) + i(y - 2) | - | (x - 3) + i(y - 4) | = 2
Simplifying this equation, we get:
√((x - 1)² + (y - 2)²) - √((x - 3)² + (y - 4)²) = 2
Squaring both sides of the equation and simplifying, we get:
(x - 1)² + (y - 2)² - 2√((x - 1)² + (y - 2)²)√((x - 3)² + (y - 4)²) + (x - 3)² + (y - 4)² = 4
Expanding and combining like terms, we get:
2x² + 2y² - 8x - 12y + 14 = 2√((x - 1)² + (y - 2)²)√((x - 3)² + (y - 4)²)
Now, we can compare this equation with the general equation of a tangent to a conic:
zz̅ - α(z + z̅) + c = 0
By comparing coefficients, we can equate the real and imaginary parts separately to find α.
From the real part: 2x² + 2y² - 8x - 12y + 14 = 0
From the imaginary part: -2{x(x - 4) + y(y - 6)} = 0
Simplifying both equations, we get:
x² - 4x + y² - 6y + 2 = 0
This is the equation of the tangent at any point (x, y) on the hyperbola.
Now, let's find the coordinates of the feet of perpendiculars from points (1, 2) and (3, 4) on this tangent.
The feet of perpendiculars can be found by substituting the coordinates of the points into the equation of the tangent and solving for x and y.
For point (
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Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer?
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Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer?.
Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer?.
Solutions for Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 isa)| z - (2 + 3i) | = 1b)| z - (2 + 3i) | = 4c)| z - (1 + i) | = 2d)| z - (1 + i) | = 1Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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