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Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve
- a)(x + y)2 = 3xy
- b)x2/3 + y2/3 = 24/3
- c)x2 + y2 = 2xy
- d)x2 + y2 = x2y2
Correct answer is option 'D'. Can you explain this answer?
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Let P be a point on the circle S with both coordinates being positive....
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Let P be a point on the circle S with both coordinates being positive....
Explanation:
We are given a circle S with a point P on it, and the tangent to the circle at point P intersects the coordinate axes at points M and N. We need to determine the curve on which the midpoint of MN lies.
Step 1: Finding the coordinates of points M and N
Since the tangent to the circle at point P is perpendicular to the radius at that point, the tangent will be parallel to the coordinate axes. Therefore, the tangent will intersect the x-axis at point M and the y-axis at point N.
Let the coordinates of point P be (a, b). Since point P lies on the circle, we have the equation of the circle as:
(x - a)^2 + (y - b)^2 = r^2
where r is the radius of the circle.
Step 2: Finding the equation of the tangent at point P
The equation of the tangent to the circle at point P can be found by differentiating the equation of the circle with respect to x and y and substituting the coordinates of point P.
Differentiating the equation of the circle with respect to x:
2(x - a) + 2(y - b) * dy/dx = 0
dy/dx = (a - x) / (y - b)
Substituting the coordinates of point P (a, b):
dy/dx = (a - a) / (b - b) = 0
Therefore, the equation of the tangent at point P is y = b.
Step 3: Finding the coordinates of points M and N
Since the tangent is parallel to the x-axis, the y-coordinate of point M is the same as the y-coordinate of point P, which is b. Therefore, point M has coordinates (x, b).
Similarly, since the tangent is parallel to the y-axis, the x-coordinate of point N is the same as the x-coordinate of point P, which is a. Therefore, point N has coordinates (a, y).
Step 4: Finding the midpoint of MN
The midpoint of a line segment with coordinates (x1, y1) and (x2, y2) can be found using the midpoint formula:
Midpoint coordinates = ((x1 + x2) / 2, (y1 + y2) / 2)
Applying this formula to points M(x, b) and N(a, y):
Midpoint coordinates = ((x + a) / 2, (b + y) / 2)
Step 5: Determining the curve on which the midpoint lies
To find the equation of the curve on which the midpoint of MN lies, we substitute the coordinates of the midpoint into the options given.
a) (x^2 + y^2)^2 = 3xy
b) (x^2/3 + y^2/3) = 24/3
c) x^2 + y^2 = 2xy
d) x^2 - y^2 = x^2y^2
Substituting the midpoint coordinates ((x + a) / 2, (b + y) / 2) into each option:
a) (((x + a) / 2)^2 + ((b + y) / 2)^2)^
We are given a circle S with a point P on it, and the tangent to the circle at point P intersects the coordinate axes at points M and N. We need to determine the curve on which the midpoint of MN lies.
Step 1: Finding the coordinates of points M and N
Since the tangent to the circle at point P is perpendicular to the radius at that point, the tangent will be parallel to the coordinate axes. Therefore, the tangent will intersect the x-axis at point M and the y-axis at point N.
Let the coordinates of point P be (a, b). Since point P lies on the circle, we have the equation of the circle as:
(x - a)^2 + (y - b)^2 = r^2
where r is the radius of the circle.
Step 2: Finding the equation of the tangent at point P
The equation of the tangent to the circle at point P can be found by differentiating the equation of the circle with respect to x and y and substituting the coordinates of point P.
Differentiating the equation of the circle with respect to x:
2(x - a) + 2(y - b) * dy/dx = 0
dy/dx = (a - x) / (y - b)
Substituting the coordinates of point P (a, b):
dy/dx = (a - a) / (b - b) = 0
Therefore, the equation of the tangent at point P is y = b.
Step 3: Finding the coordinates of points M and N
Since the tangent is parallel to the x-axis, the y-coordinate of point M is the same as the y-coordinate of point P, which is b. Therefore, point M has coordinates (x, b).
Similarly, since the tangent is parallel to the y-axis, the x-coordinate of point N is the same as the x-coordinate of point P, which is a. Therefore, point N has coordinates (a, y).
Step 4: Finding the midpoint of MN
The midpoint of a line segment with coordinates (x1, y1) and (x2, y2) can be found using the midpoint formula:
Midpoint coordinates = ((x1 + x2) / 2, (y1 + y2) / 2)
Applying this formula to points M(x, b) and N(a, y):
Midpoint coordinates = ((x + a) / 2, (b + y) / 2)
Step 5: Determining the curve on which the midpoint lies
To find the equation of the curve on which the midpoint of MN lies, we substitute the coordinates of the midpoint into the options given.
a) (x^2 + y^2)^2 = 3xy
b) (x^2/3 + y^2/3) = 24/3
c) x^2 + y^2 = 2xy
d) x^2 - y^2 = x^2y^2
Substituting the midpoint coordinates ((x + a) / 2, (b + y) / 2) into each option:
a) (((x + a) / 2)^2 + ((b + y) / 2)^2)^
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Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curvea)(x + y)2 = 3xyb)x2/3 + y2/3 = 24/3c)x2 + y2 = 2xyd)x2 + y2 = x2y2Correct answer is option 'D'. Can you explain this answer?
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Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curvea)(x + y)2 = 3xyb)x2/3 + y2/3 = 24/3c)x2 + y2 = 2xyd)x2 + y2 = x2y2Correct 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 Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curvea)(x + y)2 = 3xyb)x2/3 + y2/3 = 24/3c)x2 + y2 = 2xyd)x2 + y2 = x2y2Correct 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 Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curvea)(x + y)2 = 3xyb)x2/3 + y2/3 = 24/3c)x2 + y2 = 2xyd)x2 + y2 = x2y2Correct answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect thecoordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curvea)(x + y)2 = 3xyb)x2/3 + y2/3 = 24/3c)x2 + y2 = 2xyd)x2 + y2 = x2y2Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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