Question

If the tangent drawn at a point $P(t)$ on the hyperbola $$ x^2 - y^2 = c^2 $$ cuts the X-axis at $T$ and the normal drawn at the same point $P$ cuts the Y-axis at $N$, then the equation of the locus of the midpoint of $TN$ is:

• A. \( \frac{c^2}{4x^2} - \frac{y^2}{c^2} = 1 \)
• B. \( \frac{x^2}{4c^2} - \frac{y^2}{c^2} = 1 \)
• C. \( \frac{x^2}{c^2} - \frac{y^2}{4c^2} = 1 \)
• D. \( x^2 + y^2 = 4c^2 \)

Hint:

To find the locus of the midpoint of a chord, express the coordinates in parametric form and eliminate the parameter.

Answer 1

The Correct Option is A

Step 1: Equation of the Tangent and Normal 
For the hyperbola: \[ x^2 - y^2 = c^2, \] the equation of the tangent at a point \( P(a\sec \theta, c\tan \theta) \) is: \[ x\sec\theta - y\tan\theta = c. \] Setting \( y = 0 \) to find the X-intercept (\( T \)): \[ x = c\cos\theta. \] The equation of the normal at \( P(a\sec \theta, c\tan \theta) \) is: \[ y = -\tan\theta(x - c\sec\theta). \] Setting \( x = 0 \) to find the Y-intercept (\( N \)): \[ y = c\sin\theta. \] 

Step 2: Midpoint of \( TN \) 
The midpoint of \( TN \) is: \[ \left( \frac{c\cos\theta + 0}{2}, \frac{0 + c\sin\theta}{2} \right) = \left( \frac{c\cos\theta}{2}, \frac{c\sin\theta}{2} \right). \] 

Step 3: Finding the Locus 
Let \( X = \frac{c\cos\theta}{2} \) and \( Y = \frac{c\sin\theta}{2} \). Using \( \cos^2\theta - \sin^2\theta = 1 \), we get: \[ \frac{c^2}{4X^2} - \frac{Y^2}{c^2} = 1. \] 

Step 4: Conclusion 
Thus, the final answer is: \[ \boxed{\frac{c^2}{4x^2} - \frac{y^2}{c^2} = 1}. \]