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If the normal to the rectangular hyperbola xy = c2 at the point `t' meets the curve again at `t1' then t3t1 has the value equal to
- a)1
- b)-1
- c)0
- d)None
Correct answer is option 'B'. Can you explain this answer?
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If the normal to the rectangular hyperbola xy = c2at the point `t'...
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If the normal to the rectangular hyperbola xy = c2at the point `t'...
To solve this problem, we need to find the value of t3t1, where t is a point on the rectangular hyperbola xy = c^2 and t1 is the point where the normal to the hyperbola at t intersects the curve again.
Given that the equation of the hyperbola is xy = c^2, we can find the slope of the tangent to the curve at a point t by differentiating the equation with respect to x:
dy/dx = -c^2/x^2
The slope of the normal to the curve at t is the negative reciprocal of the slope of the tangent:
m = -x^2/c^2
Since the normal passes through the point t, we can substitute the coordinates of t into the equation of the line to find the y-intercept:
y = -x^2/c^2 * x + b
Substituting the coordinates of t = (x, y) into the equation, we get:
y = -x^3/c^2 + b
Since the normal intersects the curve again at t1, we can substitute the coordinates of t1 = (x1, y1) into the equation of the curve:
x1 * y1 = c^2
Substituting the equation of the line into the equation of the curve, we get:
x1 * (-x1^3/c^2 + b) = c^2
Simplifying the equation:
-bx1^4 + c^2x1 - c^4 = 0
This is a quartic equation in terms of x1. To find the value of x1, we can use the fact that the sum of the roots of a quartic equation is zero. Since t and t1 are two points on the curve, their x-coordinates must satisfy the equation:
x + x1 = 0
Solving for x1, we get:
x1 = -x
Therefore, the value of t3t1 is equal to -1, which corresponds to option B.
Given that the equation of the hyperbola is xy = c^2, we can find the slope of the tangent to the curve at a point t by differentiating the equation with respect to x:
dy/dx = -c^2/x^2
The slope of the normal to the curve at t is the negative reciprocal of the slope of the tangent:
m = -x^2/c^2
Since the normal passes through the point t, we can substitute the coordinates of t into the equation of the line to find the y-intercept:
y = -x^2/c^2 * x + b
Substituting the coordinates of t = (x, y) into the equation, we get:
y = -x^3/c^2 + b
Since the normal intersects the curve again at t1, we can substitute the coordinates of t1 = (x1, y1) into the equation of the curve:
x1 * y1 = c^2
Substituting the equation of the line into the equation of the curve, we get:
x1 * (-x1^3/c^2 + b) = c^2
Simplifying the equation:
-bx1^4 + c^2x1 - c^4 = 0
This is a quartic equation in terms of x1. To find the value of x1, we can use the fact that the sum of the roots of a quartic equation is zero. Since t and t1 are two points on the curve, their x-coordinates must satisfy the equation:
x + x1 = 0
Solving for x1, we get:
x1 = -x
Therefore, the value of t3t1 is equal to -1, which corresponds to option B.
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