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Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β), β > 0 is a point on this ellipse, then the equation of the normal to it at P is
- a)7c - 4y = 1
- b)4x - 2y = 1
- c)4x - 3y = 2
- d)8x - 2y = 5
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Let x = 4 be a directrix to an ellipse whose centre is at the origin a...
If the directrix of the ellipse is given by x = 4, we know that the distance between the focus (F) and any point on the ellipse (P) is equal to the distance between P and the directrix.
The distance between F and any point P on the ellipse is given by the formula:
√((x - h)^2 + (y - k)^2) = e * (x - d)
Where (h, k) is the center of the ellipse, e is the eccentricity, and d is the distance between the center and the directrix.
In this case, the center of the ellipse is at the origin (0, 0), the eccentricity is 1/2, and the directrix is x = 4.
Using the formula, we have:
√((1 - 0)^2 + (y - 0)^2) = (1/2) * (1 - 4)
Simplifying, we get:
√(1 + y^2) = (-3/2)
Squaring both sides, we have:
1 + y^2 = 9/4
Subtracting 1 from both sides, we get:
y^2 = 5/4
Taking the square root of both sides, we have:
y = ±√(5)/2
Therefore, the point P(1, ±√(5)/2) lies on the ellipse.
The distance between F and any point P on the ellipse is given by the formula:
√((x - h)^2 + (y - k)^2) = e * (x - d)
Where (h, k) is the center of the ellipse, e is the eccentricity, and d is the distance between the center and the directrix.
In this case, the center of the ellipse is at the origin (0, 0), the eccentricity is 1/2, and the directrix is x = 4.
Using the formula, we have:
√((1 - 0)^2 + (y - 0)^2) = (1/2) * (1 - 4)
Simplifying, we get:
√(1 + y^2) = (-3/2)
Squaring both sides, we have:
1 + y^2 = 9/4
Subtracting 1 from both sides, we get:
y^2 = 5/4
Taking the square root of both sides, we have:
y = ±√(5)/2
Therefore, the point P(1, ±√(5)/2) lies on the ellipse.
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Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer?
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Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. 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 x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. 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 x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer?.
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. 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 x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. 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 x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer?.
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Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 1/2. if P(1,β),β > 0is a point on this ellipse, then the equation of the normal to it at P isa)7c - 4y = 1b)4x - 2y = 1c)4x - 3y = 2d)8x - 2y = 5Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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