JEE Exam > JEE Questions > Let P be a variable point on the parabola y =...
Start Learning for Free
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:
- a)(3x - y)2 + (x - 3y) + 2 = 0
- b)2(x - 3y)2 + (3x - y) + 2 = 0
- c)2(3x - y)2 + (x - 3y) + 2 = 0
- d)(3x - y)2 + 2(x - 3y) + 2 = 0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus ...
Explanation:
Given, the equation of parabola is y = 4x^2 - 1.
Let P be a variable point on the parabola and Q be the foot of the perpendicular drawn from P to the line y = x.
Let R be the mid-point of P and Q.
Let the coordinates of P be (a, 4a^2 - 1).
The slope of the line PQ is -1.
Therefore, the equation of PQ is y - (4a^2 - 1) = -1(x - a).
Simplifying the above equation, we get x + y = 4a^2.
The equation of the tangent to the parabola at P is y = 8ax - 4a^2 - 1.
The slope of the tangent at P is 8a.
Therefore, the slope of the normal at P is -1/8a.
The equation of the normal at P is y - (4a^2 - 1) = (-1/8a)(x - a).
Simplifying the above equation, we get x + 8ay = 32a^3 - 1.
The coordinates of Q can be found by solving the equations of PQ and the normal at P.
On solving, we get Q as (16a^3 - a, 16a^3 - 4a^2).
The coordinates of R, the mid-point of P and Q, can be found as ((a + 16a^3 - a)/2, (4a^2 - 1 + 16a^3 - 4a^2)/2).
Simplifying the above equation, we get R as (8a^3, 8a^3 - 1/2).
Eliminating a from the equations of x and y of R, we get the locus of R as:
2(3x - y)^2 * (x - 3y) = 0.
Therefore, the correct option is (c).
Given, the equation of parabola is y = 4x^2 - 1.
Let P be a variable point on the parabola and Q be the foot of the perpendicular drawn from P to the line y = x.
Let R be the mid-point of P and Q.
Let the coordinates of P be (a, 4a^2 - 1).
The slope of the line PQ is -1.
Therefore, the equation of PQ is y - (4a^2 - 1) = -1(x - a).
Simplifying the above equation, we get x + y = 4a^2.
The equation of the tangent to the parabola at P is y = 8ax - 4a^2 - 1.
The slope of the tangent at P is 8a.
Therefore, the slope of the normal at P is -1/8a.
The equation of the normal at P is y - (4a^2 - 1) = (-1/8a)(x - a).
Simplifying the above equation, we get x + 8ay = 32a^3 - 1.
The coordinates of Q can be found by solving the equations of PQ and the normal at P.
On solving, we get Q as (16a^3 - a, 16a^3 - 4a^2).
The coordinates of R, the mid-point of P and Q, can be found as ((a + 16a^3 - a)/2, (4a^2 - 1 + 16a^3 - 4a^2)/2).
Simplifying the above equation, we get R as (8a^3, 8a^3 - 1/2).
Eliminating a from the equations of x and y of R, we get the locus of R as:
2(3x - y)^2 * (x - 3y) = 0.
Therefore, the correct option is (c).
Free Test
FREE
| Start Free Test |
Community Answer
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus ...
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer?
Question Description
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. 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 Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer?.
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. 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 Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer?.
Solutions for Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let P be a variable point on the parabola y = 4x2 + 1. Then the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is:a)(3x - y)2 + (x - 3y) + 2 = 0b)2(x - 3y)2 + (3x - y) + 2 = 0c)2(3x - y)2 + (x - 3y) + 2 = 0d)(3x - y)2 + 2(x - 3y) + 2 = 0Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup