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PQ is a double ordinate of the ellipse x2+ 9y2 = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is
- a)A circle
- b)A parabola
- c)An ellipse
- d)A hyperbola
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P mee...
Equation of ellipse : x2/9 + y2/1= 1
Co−ordinates of point P is : (acosθ , bsinθ)
Equation of normal at point (x1, y1) is :
a2x/x1 − b2y/y1 = (ae)2
Equation of normal at point P is:
ax/cosθ− by/sinθ= (ae)2−−−−−(1)
Equation of diameter at Q is : y = (−b/a)tanθ−−−−−(2)
Point of intersection of equation 1 and 2 is :
ax/cosθ + (b2/acosθ) = (ae)2
Or , x = ae2cosθ − b2/a2
And y = − be2sinθ + b3/a3 (from 2)
∴ Point R(ae2cosθ − b2/a2 , − be2sinθ + b3/a3)
Let mid point of PR is (h,k)
h =[acosθ + ae2cosθ − (b2/a2)]/2
Or , 2h + b2/a2= acosθ + ae2cosθ
Or , cosθ= [2h + (b2/a2)/(a + ae2)]
k = [bsinθ − be2sinθ + (b3/a3/2)]/2
Or , sinθ = [2k − (b3/a3)/(b − be2)]
On squaring adding we get :
[2h + (b2/a2)]/(a + ae2)2 + [2k − (b3/a3)] / (b − be2)2= 1
Which is an equation of ellipse.
Co−ordinates of point P is : (acosθ , bsinθ)
Equation of normal at point (x1, y1) is :
a2x/x1 − b2y/y1 = (ae)2
Equation of normal at point P is:
ax/cosθ− by/sinθ= (ae)2−−−−−(1)
Equation of diameter at Q is : y = (−b/a)tanθ−−−−−(2)
Point of intersection of equation 1 and 2 is :
ax/cosθ + (b2/acosθ) = (ae)2
Or , x = ae2cosθ − b2/a2
And y = − be2sinθ + b3/a3 (from 2)
∴ Point R(ae2cosθ − b2/a2 , − be2sinθ + b3/a3)
Let mid point of PR is (h,k)
h =[acosθ + ae2cosθ − (b2/a2)]/2
Or , 2h + b2/a2= acosθ + ae2cosθ
Or , cosθ= [2h + (b2/a2)/(a + ae2)]
k = [bsinθ − be2sinθ + (b3/a3/2)]/2
Or , sinθ = [2k − (b3/a3)/(b − be2)]
On squaring adding we get :
[2h + (b2/a2)]/(a + ae2)2 + [2k − (b3/a3)] / (b − be2)2= 1
Which is an equation of ellipse.
Most Upvoted Answer
PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P mee...
Given an ellipse with equation x^2/9 + y^2/1 = 1, let's analyze the given information step by step to understand why the locus of the midpoint of PR is an ellipse.
1. Finding the coordinates of point P:
Since PQ is a double ordinate, it means that the line segment PQ passes through the center of the ellipse. The center of the ellipse is (0, 0), so the coordinates of point P are (0, 1).
2. Finding the equation of the normal at point P:
To find the equation of the normal at point P, we need to find the slope of the tangent at point P. Differentiating the equation of the ellipse, we get:
2x/9 + 2y(dy/dx) = 0
dy/dx = -9x/(2y)
The slope of the normal is the negative reciprocal of the slope of the tangent, so the slope of the normal at P is (2y)/(9x).
Using the point-slope form of a line, the equation of the normal at P is given by:
y - 1 = (2y)/(9x) * (x - 0)
9xy - 9x = 2y^2 - 2y
2y^2 - 9xy + 9x - 2y + 2 = 0
3. Finding the coordinates of point Q:
Since PQ is a double ordinate, it means that the distance between points P and Q is equal to 2 times the distance between point P and the center of the ellipse. The distance between P and the center is 1, so the distance between P and Q is 2.
Since point P is at (0, 1), point Q can be at either (2, 1) or (-2, 1).
4. Finding the equation of the diameter through point Q:
Since point Q lies on the diameter, we can find the equation of the diameter passing through point Q by finding the equation of the line passing through points P and Q.
Using the two-point form of a line, the equation of the diameter is given by:
(y - 1) = (1 - 1)/(0 - 2) * (x - 2)
(x + 2) = 0
So the equation of the diameter is x = -2.
5. Finding the coordinates of point R:
Point R is the intersection of the normal at P and the diameter passing through Q. Since the equation of the diameter is x = -2, the x-coordinate of point R is -2. We can substitute this value into the equation of the normal to find the y-coordinate of point R.
2y^2 - 9(-2)y + 9(-2) - 2y + 2 = 0
2y^2 + 18y - 18 - 2y + 2 = 0
2y^2 + 16y - 16 = 0
y^2 + 8y - 8 = 0
Using the quadratic formula, we can solve for y:
y = (-8 ± √(8^2 - 4*1*(-8)))/2
y = (-8 ± √(64 + 32))/2
y = (-8 ± √96)/2
y = (-
1. Finding the coordinates of point P:
Since PQ is a double ordinate, it means that the line segment PQ passes through the center of the ellipse. The center of the ellipse is (0, 0), so the coordinates of point P are (0, 1).
2. Finding the equation of the normal at point P:
To find the equation of the normal at point P, we need to find the slope of the tangent at point P. Differentiating the equation of the ellipse, we get:
2x/9 + 2y(dy/dx) = 0
dy/dx = -9x/(2y)
The slope of the normal is the negative reciprocal of the slope of the tangent, so the slope of the normal at P is (2y)/(9x).
Using the point-slope form of a line, the equation of the normal at P is given by:
y - 1 = (2y)/(9x) * (x - 0)
9xy - 9x = 2y^2 - 2y
2y^2 - 9xy + 9x - 2y + 2 = 0
3. Finding the coordinates of point Q:
Since PQ is a double ordinate, it means that the distance between points P and Q is equal to 2 times the distance between point P and the center of the ellipse. The distance between P and the center is 1, so the distance between P and Q is 2.
Since point P is at (0, 1), point Q can be at either (2, 1) or (-2, 1).
4. Finding the equation of the diameter through point Q:
Since point Q lies on the diameter, we can find the equation of the diameter passing through point Q by finding the equation of the line passing through points P and Q.
Using the two-point form of a line, the equation of the diameter is given by:
(y - 1) = (1 - 1)/(0 - 2) * (x - 2)
(x + 2) = 0
So the equation of the diameter is x = -2.
5. Finding the coordinates of point R:
Point R is the intersection of the normal at P and the diameter passing through Q. Since the equation of the diameter is x = -2, the x-coordinate of point R is -2. We can substitute this value into the equation of the normal to find the y-coordinate of point R.
2y^2 - 9(-2)y + 9(-2) - 2y + 2 = 0
2y^2 + 18y - 18 - 2y + 2 = 0
2y^2 + 16y - 16 = 0
y^2 + 8y - 8 = 0
Using the quadratic formula, we can solve for y:
y = (-8 ± √(8^2 - 4*1*(-8)))/2
y = (-8 ± √(64 + 32))/2
y = (-8 ± √96)/2
y = (-
Community Answer
PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P mee...
Option C is correct...
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PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer?
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PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. 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 PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer?.
PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. 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 PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer?.
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PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of PQ is a double ordinate of the ellipse x2+ 9y2= 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR isa)A circleb)A parabolac)An ellipsed)A hyperbolaCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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