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Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 is
- a)y + mx = 0
- b)y – mx = 0
- c)my – mx = 0
- d)my + x = 0
Correct answer is option 'A'. Can you explain this answer?
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Locus of the middle points of the parallel chords with gradient m of t...
Let the mid point of chord be p (h,k) The equation of chord with p as mid point is Kx + hy =2hk The given slope m of the chord is m Hence -k/h =m Therefore required ans is y+mx =0 .
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Locus of the middle points of the parallel chords with gradient m of t...
The rectangular hyperbola xy = c^2 can be written as y = c^2/x.
a) Let's first consider the case where the gradient of the parallel chords is m and the chord passes through the point (x1, y1).
The equation of the chord can be written as y - y1 = m(x - x1). Substituting y = c^2/x, we get:
c^2/x - y1 = m(x - x1)
Rearranging and simplifying, we get:
x^2 - (m+ y1)x + c^2/m = 0
This is a quadratic equation in x, and we know that the midpoint of the chord lies on the line x = (x1 + x2)/2, where (x2, y2) is the other point on the chord. Therefore, the x-coordinate of the midpoint is:
(x1 + x2)/2 = (m + y1)/2
Solving for x2 using the quadratic equation above and substituting into the expression for the x-coordinate of the midpoint, we get:
x = (m + y1)/2 +- c/sqrt(m^2 + 1)
Therefore, the locus of the midpoints of the parallel chords with gradient m is a pair of straight lines given by the above equation.
b) Now let's consider the case where the chord is parallel to the y-axis, i.e., m = infinity. In this case, the equation of the chord is simply x = x1, and substituting into the equation of the hyperbola, we get y = c^2/x1. Therefore, the midpoint of the chord is (x1, c^2/x1), which lies on the hyperbola y = c^2/x.
Therefore, the locus of the midpoints of the parallel chords with gradient infinity is the hyperbola y = c^2/x.
a) Let's first consider the case where the gradient of the parallel chords is m and the chord passes through the point (x1, y1).
The equation of the chord can be written as y - y1 = m(x - x1). Substituting y = c^2/x, we get:
c^2/x - y1 = m(x - x1)
Rearranging and simplifying, we get:
x^2 - (m+ y1)x + c^2/m = 0
This is a quadratic equation in x, and we know that the midpoint of the chord lies on the line x = (x1 + x2)/2, where (x2, y2) is the other point on the chord. Therefore, the x-coordinate of the midpoint is:
(x1 + x2)/2 = (m + y1)/2
Solving for x2 using the quadratic equation above and substituting into the expression for the x-coordinate of the midpoint, we get:
x = (m + y1)/2 +- c/sqrt(m^2 + 1)
Therefore, the locus of the midpoints of the parallel chords with gradient m is a pair of straight lines given by the above equation.
b) Now let's consider the case where the chord is parallel to the y-axis, i.e., m = infinity. In this case, the equation of the chord is simply x = x1, and substituting into the equation of the hyperbola, we get y = c^2/x1. Therefore, the midpoint of the chord is (x1, c^2/x1), which lies on the hyperbola y = c^2/x.
Therefore, the locus of the midpoints of the parallel chords with gradient infinity is the hyperbola y = c^2/x.
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Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2isa)y + mx = 0b)y– mx = 0c)my –mx = 0d)my + x = 0Correct answer is option 'A'. Can you explain this answer?
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