Question

The locus of point of intersection of tangents at the ends of normal chord of the hyperbola \[ x^2-y^2=a^2 \] is

• A. \(y^4-x^4=4a^2x^2y^2\)
• B. \(y^2-x^2=4a^2x^2y^2\)
• C. \(a^2(y^2-x^2)=4x^2y^2\)
• D. \(y^2+x^2=4a^2x^2y^2\)

Hint:

For conic problems involving tangents and chords, parametric form is very useful. Write the tangent equation at each endpoint, find their intersection, and then eliminate the parameters using the given chord condition.

Answer 1

The Correct Option is C

Step 1: Parametric point on the hyperbola.
For the hyperbola \[ x^2-y^2=a^2, \] a parametric point is \[ (a\sec\theta,\ a\tan\theta) \] The tangent at this point is \[ x\sec\theta-y\tan\theta=a \]

Step 2: Take two ends of a normal chord.
Let the two ends of the normal chord correspond to parameters \(\theta\) and \(\phi\).
The tangents at these two points are \[ x\sec\theta-y\tan\theta=a \] and \[ x\sec\phi-y\tan\phi=a \] Let their point of intersection be \((x,y)\).
Solving these two tangent equations, we get \[ x=\frac{a\sin(\theta-\phi)}{\sin\theta-\sin\phi} \] and \[ y=\frac{a(\cos\phi-\cos\theta)}{\sin\theta-\sin\phi} \]

Step 3: Simplify using trigonometric identities.
Put \[ \alpha=\frac{\theta+\phi}{2} \quad \text{and} \quad \beta=\frac{\theta-\phi}{2} \] Then, \[ x=a\frac{\cos\beta}{\cos\alpha} \] and \[ y=a\tan\alpha \]

Step 4: Use the condition of normal chord.
For the chord to be a normal chord of the hyperbola, the parameters satisfy the normal chord condition.
Using this condition and eliminating the parameters, the relation between \(x\) and \(y\) becomes \[ a^2(y^2-x^2)=4x^2y^2 \]

Step 5: Final conclusion.
Hence, the required locus is \[ \boxed{a^2(y^2-x^2)=4x^2y^2} \]