JEE Exam > JEE Questions > Let the abscissae of the two points P and Q b...
Start Learning for Free
Let the abscissae of the two points P and Q be the roots of 2x2 - rx + p = 0 and the ordinates of P and Q be the roots of x2 - sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2 + y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal to
Correct answer is '7'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Let the abscissae of the two points P and Q be the roots of 2x2- rx + ...
Equation of the circle with PQ as diameter is 2(x2 + y2) - rx - 2sy + p - 2q = 0
On comparing with the given equation,
r = 11, s = 7
p - 2q = -22
∴ 2r + s - 2q + p = 22 + 7 - 22 = 7
Free Test
FREE
| Start Free Test |
Community Answer
Let the abscissae of the two points P and Q be the roots of 2x2- rx + ...
Given information:
The equation of the circle described on PQ as diameter is 2(x^2 + y^2) - 11x - 14y - 22 = 0.
Step 1: Finding the coordinates of P and Q
The abscissae of P and Q are the roots of the equation 2x^2 - rx + p = 0.
Similarly, the ordinates of P and Q are the roots of the equation x^2 - sx - q = 0.
Let's solve these equations to find the coordinates of P and Q.
Step 2: Finding the abscissae of P and Q
The equation 2x^2 - rx + p = 0 has roots equal to the abscissae of P and Q.
Let the roots be α and β.
By Vieta's formulas, we know that α + β = r/2 and αβ = p/2.
Step 3: Finding the ordinates of P and Q
The equation x^2 - sx - q = 0 has roots equal to the ordinates of P and Q.
Let the roots be γ and δ.
By Vieta's formulas, we know that γ + δ = s/2 and γδ = -q.
Step 4: Finding the coordinates of P and Q
The coordinates of P are (α, γ), and the coordinates of Q are (β, δ).
Step 5: Using the equation of the circle
We know that the circle described on PQ as diameter has the equation 2(x^2 + y^2) - 11x - 14y - 22 = 0.
Step 6: Substituting the coordinates of P and Q in the equation of the circle
Substituting the coordinates of P and Q in the equation of the circle, we get:
2(α^2 + γ^2) - 11α - 14γ - 22 = 0
2(β^2 + δ^2) - 11β - 14δ - 22 = 0
Step 7: Simplifying the equations
Expanding the expressions and collecting like terms, we get:
2α^2 + 2γ^2 - 11α - 14γ - 22 = 0
2β^2 + 2δ^2 - 11β - 14δ - 22 = 0
Step 8: Manipulating the equations
Adding the two equations, we get:
2(α^2 + β^2 + γ^2 + δ^2) - 11(α + β) - 14(γ + δ) - 44 = 0
Step 9: Simplifying further using Vieta's formulas
Using Vieta's formulas, we can substitute the values of α + β and γ + δ:
2(α^2 + β^2 + γ^2 + δ^2) - 11(r/2) - 14(s/2) - 44 = 0
2(α^2 + β^2 + γ^2 + δ^2) - 11r/2 -
The equation of the circle described on PQ as diameter is 2(x^2 + y^2) - 11x - 14y - 22 = 0.
Step 1: Finding the coordinates of P and Q
The abscissae of P and Q are the roots of the equation 2x^2 - rx + p = 0.
Similarly, the ordinates of P and Q are the roots of the equation x^2 - sx - q = 0.
Let's solve these equations to find the coordinates of P and Q.
Step 2: Finding the abscissae of P and Q
The equation 2x^2 - rx + p = 0 has roots equal to the abscissae of P and Q.
Let the roots be α and β.
By Vieta's formulas, we know that α + β = r/2 and αβ = p/2.
Step 3: Finding the ordinates of P and Q
The equation x^2 - sx - q = 0 has roots equal to the ordinates of P and Q.
Let the roots be γ and δ.
By Vieta's formulas, we know that γ + δ = s/2 and γδ = -q.
Step 4: Finding the coordinates of P and Q
The coordinates of P are (α, γ), and the coordinates of Q are (β, δ).
Step 5: Using the equation of the circle
We know that the circle described on PQ as diameter has the equation 2(x^2 + y^2) - 11x - 14y - 22 = 0.
Step 6: Substituting the coordinates of P and Q in the equation of the circle
Substituting the coordinates of P and Q in the equation of the circle, we get:
2(α^2 + γ^2) - 11α - 14γ - 22 = 0
2(β^2 + δ^2) - 11β - 14δ - 22 = 0
Step 7: Simplifying the equations
Expanding the expressions and collecting like terms, we get:
2α^2 + 2γ^2 - 11α - 14γ - 22 = 0
2β^2 + 2δ^2 - 11β - 14δ - 22 = 0
Step 8: Manipulating the equations
Adding the two equations, we get:
2(α^2 + β^2 + γ^2 + δ^2) - 11(α + β) - 14(γ + δ) - 44 = 0
Step 9: Simplifying further using Vieta's formulas
Using Vieta's formulas, we can substitute the values of α + β and γ + δ:
2(α^2 + β^2 + γ^2 + δ^2) - 11(r/2) - 14(s/2) - 44 = 0
2(α^2 + β^2 + γ^2 + δ^2) - 11r/2 -
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer?
Question Description
Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. 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 Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer?.
Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. 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 Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer?.
Solutions for Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. 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 the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer?, a detailed solution for Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? has been provided alongside types of Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let the abscissae of the two points P and Q be the roots of 2x2- rx + p = 0 and the ordinates of P and Q be the roots of x2- sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x2+ y2) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal toCorrect answer is '7'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
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