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The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is
- a)(x2 - y2)2 = 6x2 + 2y2
- b)(x2 - y2)2 = 6x2 - 2y2
- c)(x2 + y2)2 = 6x2 + 2y2
- d)(x + y)2 = 6x2 + 2y2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The locus of the foot of perpendicular drawn from the centre of the el...
Locus of the foot of perpendicular drawn from the centre of the ellipse x²/6 + y²/2 = 1 on any tangent to it is given by the equation (x² - y²)² = 6x² - 2y².
Explanation:
The equation of the ellipse is given as x²/6 + y²/2 = 1, which can be rewritten as:
x²/6 = 1 - y²/2
Taking the derivative of both sides with respect to x, we get:
d/dx(x²/6) = d/dx(1 - y²/2)
2x/6 = 0 - (-2y/2) * dy/dx
Simplifying the above equation, we get:
x/3 = y * dy/dx
The derivative dy/dx represents the slope of the tangent to the ellipse at any point (x, y). The perpendicular drawn from the centre of the ellipse to the tangent line has a slope equal to the negative reciprocal of the tangent line's slope.
Therefore, the slope of the perpendicular line is -3/y (since the slope of the tangent line is y * dy/dx).
Using the point-slope form of a line, the equation of the perpendicular line passing through the centre (0, 0) is given by:
y - 0 = (-3/y)(x - 0)
Simplifying the equation, we have:
y = (-3/y)x
Multiplying both sides by y, we get:
y² = -3x
Substituting this value of y² into the equation x²/6 + y²/2 = 1, we have:
x²/6 + (-3x)/2 = 1
Multiplying through by 6 to eliminate the fractions, we get:
x² - 9x = 6
Rearranging the terms, we have:
x² - 9x - 6 = 0
Now, let's solve this quadratic equation to find the values of x. Using the quadratic formula, we have:
x = (-(-9) ± √((-9)² - 4(1)(-6))) / (2(1))
Simplifying further, we get:
x = (9 ± √(81 + 24)) / 2
x = (9 ± √105) / 2
Therefore, the locus of the foot of perpendicular drawn from the centre is given by the equation:
(x² - y²)² = 6x² - 2y²
Hence, the correct answer is option 'C'.
Explanation:
The equation of the ellipse is given as x²/6 + y²/2 = 1, which can be rewritten as:
x²/6 = 1 - y²/2
Taking the derivative of both sides with respect to x, we get:
d/dx(x²/6) = d/dx(1 - y²/2)
2x/6 = 0 - (-2y/2) * dy/dx
Simplifying the above equation, we get:
x/3 = y * dy/dx
The derivative dy/dx represents the slope of the tangent to the ellipse at any point (x, y). The perpendicular drawn from the centre of the ellipse to the tangent line has a slope equal to the negative reciprocal of the tangent line's slope.
Therefore, the slope of the perpendicular line is -3/y (since the slope of the tangent line is y * dy/dx).
Using the point-slope form of a line, the equation of the perpendicular line passing through the centre (0, 0) is given by:
y - 0 = (-3/y)(x - 0)
Simplifying the equation, we have:
y = (-3/y)x
Multiplying both sides by y, we get:
y² = -3x
Substituting this value of y² into the equation x²/6 + y²/2 = 1, we have:
x²/6 + (-3x)/2 = 1
Multiplying through by 6 to eliminate the fractions, we get:
x² - 9x = 6
Rearranging the terms, we have:
x² - 9x - 6 = 0
Now, let's solve this quadratic equation to find the values of x. Using the quadratic formula, we have:
x = (-(-9) ± √((-9)² - 4(1)(-6))) / (2(1))
Simplifying further, we get:
x = (9 ± √(81 + 24)) / 2
x = (9 ± √105) / 2
Therefore, the locus of the foot of perpendicular drawn from the centre is given by the equation:
(x² - y²)² = 6x² - 2y²
Hence, the correct answer is option 'C'.
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Community Answer
The locus of the foot of perpendicular drawn from the centre of the el...
Let the foot of perpendicular be P(h, k)
Equation of tangent with slope m passing P(h, k) is
So required locus is (x2 + y2)2 = 6x2 + 2y2
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The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer?
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The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. 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 locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. 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 locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer?.
The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. 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 locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. 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 locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer?.
Solutions for The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it isa)(x2 - y2)2 = 6x2 + 2y2b)(x2 - y2)2 = 6x2 - 2y2c)(x2 + y2)2 = 6x2 + 2y2d)(x + y)2 = 6x2 + 2y2Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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