The locus of the midpoint of the chords of the hyperbola `(x^(2))/(25)-(y^(2))/(36)=1` which passes through the point (2, 4) is a hyperbola, whose transverse axis length (in units) is equal to

Options

A. `(16)/(5)`
B. `(4)/(3)`
C. `(8)/(5)`
D. `(61)/(25)`

Answer

To solve the problem step by step, we need to find the locus of the midpoint of the chords of the hyperbola \(\frac{x^2}{25} - \frac{y^2}{36} = 1\) that passes through the point (2, 4).

### Step 1: Identify the hyperbola
The given hyperbola is \(\frac{x^2}{25} - \frac{y^2}{36} = 1\). Here, \(a^2 = 25\) and \(b^2 = 36\). This means \(a = 5\) and \(b = 6\).

### Step 2: Find the coordinates of points on the hyperbola
Let points \(A\) and \(B\) be two points on the hyperbola. We can represent these points in parametric form:
- \(A = (5 \sec \theta_1, 6 \tan \theta_1)\)
- \(B = (5 \sec \theta_2, 6 \tan \theta_2)\)

### Step 3: Find the midpoint of the chord
The midpoint \(M(h, k)\) of the chord \(AB\) is given by:
\[ h = \frac{5 \sec \theta_1 + 5 \sec \theta_2}{2} = \frac{5}{2} (\sec \theta_1 + \sec \theta_2) \]
\[ k = \frac{6 \tan \theta_1 + 6 \tan \theta_2}{2} = 3 (\tan \theta_1 + \tan \theta_2) \]

### Step 4: Express \(h\) and \(k\) in terms of trigonometric identities
Using the identities:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
We can express \(h^2\) and \(k^2\):
\[ h^2 = \left(\frac{5}{2} (\sec \theta_1 + \sec \theta_2)\right)^2 = \frac{25}{4} (\sec^2 \theta_1 + \sec^2 \theta_2 + 2 \sec \theta_1 \sec \theta_2) \]
\[ k^2 = \left(3 (\tan \theta_1 + \tan \theta_2)\right)^2 = 9 (\tan^2 \theta_1 + \tan^2 \theta_2 + 2 \tan \theta_1 \tan \theta_2) \]

### Step 5: Set up the equation for the locus
Using the relationship between secant and tangent, we can substitute and simplify to find the equation of the locus of the midpoint:
\[ \frac{4h^2}{25} - \frac{k^2}{9} = \lambda \]
where \(\lambda\) is a constant.

### Step 6: Substitute the point (2, 4)
Since the locus passes through the point (2, 4):
\[ \frac{4(2)^2}{25} - \frac{(4)^2}{9} = \lambda \]
Calculating this gives:
\[ \frac{16}{25} - \frac{16}{9} = \lambda \]
Finding a common denominator and simplifying gives us the value of \(\lambda\).

### Step 7: Find the transverse axis length
The equation of the hyperbola in standard form is:
\[ \frac{h^2}{\frac{25}{2}} - \frac{k^2}{18} = 1 \]
The length of the transverse axis is given by \(2a\), where \(a = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}\). Therefore, the length of the transverse axis is:
\[ 2a = 2 \cdot \frac{5}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

### Final Answer
The length of the transverse axis is \(5\sqrt{2}\) units.

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