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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?
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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
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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 -
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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?
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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?.
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