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The line x = 2y intersects the ellipse x2/4 + y2 = 1 at the points P and Q. The equation of the circle with PQ as diameter is
- a)x2 + y2 = 1/2
- b)x2 + y2 = 1
- c)x2 + y2 = 2
- d)x2 + y2 = 5/2
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and...
Solving the given equations then common points are 
Eqn. of Circle PQ as
diameter
diameter
Most Upvoted Answer
The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and...
Given:
- The line x = 2y intersects the ellipse x^2/4 + y^2 = 1 at the points P and Q.
To find:
- The equation of the circle with PQ as diameter.
Solution:
Step 1: Find the coordinates of points P and Q:
- Substitute x = 2y in the equation of the ellipse:
(2y)^2/4 + y^2 = 1
4y^2/4 + y^2 = 1
5y^2/4 = 1
y^2 = 4/5
y = ±√(4/5)
- Since x = 2y, we can substitute the value of y in x = 2y:
x = 2(±√(4/5))
x = ±2√(4/5)
- Therefore, the coordinates of points P and Q are (2√(4/5), ±√(4/5)).
Step 2: Find the midpoint of PQ:
- The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is given by the formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
- Applying this formula, we can find the coordinates of the midpoint of PQ:
Midpoint = ((2√(4/5) + -2√(4/5))/2, (√(4/5) + -√(4/5))/2)
Midpoint = (0, 0)
Step 3: Find the radius of the circle:
- The radius of a circle with diameter PQ is the distance from the midpoint to one of the endpoints.
- Using the distance formula, we can calculate the radius:
Radius = √((x1 - x2)^2 + (y1 - y2)^2)
Radius = √((0 - 2√(4/5))^2 + (0 - √(4/5))^2)
Radius = √((-2√(4/5))^2 + (-√(4/5))^2)
Radius = √(4(4/5) + (4/5))
Radius = √(16/5 + 4/5)
Radius = √(20/5)
Radius = √4
Radius = 2
Step 4: Write the equation of the circle:
- The equation of a circle with center (h, k) and radius r is given by the formula:
(x - h)^2 + (y - k)^2 = r^2
- Since the midpoint of PQ is (0, 0) and the radius is 2, the equation of the circle is:
(x - 0)^2 + (y - 0)^2 = 2^2
x^2 + y^2 = 4
Final Answer:
The equation of the circle with PQ as diameter is x^2 + y^2 = 4, which is option D.
- The line x = 2y intersects the ellipse x^2/4 + y^2 = 1 at the points P and Q.
To find:
- The equation of the circle with PQ as diameter.
Solution:
Step 1: Find the coordinates of points P and Q:
- Substitute x = 2y in the equation of the ellipse:
(2y)^2/4 + y^2 = 1
4y^2/4 + y^2 = 1
5y^2/4 = 1
y^2 = 4/5
y = ±√(4/5)
- Since x = 2y, we can substitute the value of y in x = 2y:
x = 2(±√(4/5))
x = ±2√(4/5)
- Therefore, the coordinates of points P and Q are (2√(4/5), ±√(4/5)).
Step 2: Find the midpoint of PQ:
- The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is given by the formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
- Applying this formula, we can find the coordinates of the midpoint of PQ:
Midpoint = ((2√(4/5) + -2√(4/5))/2, (√(4/5) + -√(4/5))/2)
Midpoint = (0, 0)
Step 3: Find the radius of the circle:
- The radius of a circle with diameter PQ is the distance from the midpoint to one of the endpoints.
- Using the distance formula, we can calculate the radius:
Radius = √((x1 - x2)^2 + (y1 - y2)^2)
Radius = √((0 - 2√(4/5))^2 + (0 - √(4/5))^2)
Radius = √((-2√(4/5))^2 + (-√(4/5))^2)
Radius = √(4(4/5) + (4/5))
Radius = √(16/5 + 4/5)
Radius = √(20/5)
Radius = √4
Radius = 2
Step 4: Write the equation of the circle:
- The equation of a circle with center (h, k) and radius r is given by the formula:
(x - h)^2 + (y - k)^2 = r^2
- Since the midpoint of PQ is (0, 0) and the radius is 2, the equation of the circle is:
(x - 0)^2 + (y - 0)^2 = 2^2
x^2 + y^2 = 4
Final Answer:
The equation of the circle with PQ as diameter is x^2 + y^2 = 4, which is option D.
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The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. 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 The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. Can you explain this answer?.
The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. 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 The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The line x = 2y intersects the ellipse x2/4 + y2= 1at the points P and Q. The equation of the circle with PQ as diameter isa)x2+ y2= 1/2b)x2+ y2= 1c)x2+ y2= 2d)x2+ y2= 5/2Correct answer is option 'D'. Can you explain this answer?.
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