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The locus of the middle points of the chords of hyperbola , x²/9-y²/4=1which pass through the fixed point (1, 2) is a hyperbola with the centre .?
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The locus of the middle points of the chords of hyperbola , x²/9-y²/4=...
Given Hyperbola:
The given hyperbola equation is x²/9 - y²/4 = 1.

Definition of a Chord:
A chord of a curve is a line segment whose endpoints lie on the curve.

Definition of the Locus:
The locus of a point is the set of all points that satisfy a given condition or set of conditions.

Finding the Locus of the Middle Points of the Chords:
To find the locus of the middle points of the chords of the given hyperbola that pass through the fixed point (1, 2), we can follow these steps:

1. Consider a general point (h, k) on the hyperbola.
2. Find the equation of the chord passing through (1, 2) and (h, k).
3. Find the midpoint of this chord using the midpoint formula.
4. Express the coordinates of the midpoint in terms of h and k.
5. Substitute the equation of the hyperbola into the coordinates of the midpoint to eliminate h and k.
6. Simplify the resulting equation to obtain the locus of the middle points.

Equation of the Chord:
The equation of the chord passing through (1, 2) and (h, k) can be found using the two-point form of a line:

(y - 2) = ((k - 2)/(h - 1))(x - 1)

Simplifying this equation gives:

y = ((k - 2)/(h - 1))(x - 1) + 2

Finding the Midpoint:
The midpoint of the chord is given by the average of the x-coordinates and the average of the y-coordinates of the endpoints. Let the midpoint be (a, b), then we have:

a = (1 + h)/2
b = (2 + k)/2

Expressing the Midpoint in Terms of h and k:
From the above equations, we can express h and k in terms of a and b:

h = 2a - 1
k = 2b - 2

Substituting into the Hyperbola Equation:
Substituting h = 2a - 1 and k = 2b - 2 into the equation of the hyperbola x²/9 - y²/4 = 1, we get:

((2a - 1)²)/9 - ((2b - 2)²)/4 = 1

Simplifying this equation gives:

4a² - 4a + 9b² - 36b + 29 = 0

Simplifying the Equation:
Rearranging the terms, we get:

4a² - 4a + 9b² - 36b = -29

Dividing through by -29, we obtain:

a²/(-29/4) - a/(-29/4) + b²/(-29/9) - b/(-29/9) = 1

Comparing this equation with the standard form of a hyperbola, we have:

x²/(-29/4) - y²/(-29/9) = 1

Conclusion:
Therefore
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The locus of the middle points of the chords of hyperbola , x²/9-y²/4=1which pass through the fixed point (1, 2) is a hyperbola with the centre .?
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The locus of the middle points of the chords of hyperbola , x²/9-y²/4=1which pass through the fixed point (1, 2) is a hyperbola with the centre .? 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 middle points of the chords of hyperbola , x²/9-y²/4=1which pass through the fixed point (1, 2) is a hyperbola with the centre .? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The locus of the middle points of the chords of hyperbola , x²/9-y²/4=1which pass through the fixed point (1, 2) is a hyperbola with the centre .?.
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