Let $P(a~sec~ heta, b~tan~ heta)$ and $Q(a~sec~\phi, b~tan~\phi)$, where $ heta+\phi=\frac{\pi}{2}.$ be two points on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If (h, k) is the point of intersection of the normals of P and Q, then $k=$

Q: Let $P(a~sec~\theta, b~tan~\theta)$ and $Q(a~sec~\phi, b~tan~\phi)$, where $\theta+\phi=\frac{\pi}{2}.$ be two points on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If (h, k) is the point of intersection of the normals of P and Q, then $k=$

Let $P(a~sec~\theta, b~tan~\theta)$ and $Q(a~sec~\phi, b~tan~\phi)$, where $\theta+\phi=\frac{\pi}{2}.$ be two points on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If (h, k) is the point of intersection of the normals of P and Q, then $k=$

Question

Let $P(a~sec~	heta, b~tan~	heta)$ and $Q(a~sec~\phi, b~tan~\phi)$, where $	heta+\phi=\frac{\pi}{2}.$ be two points on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If (h, k) is the point of intersection of the normals of P and Q, then $k=$

Accepted Answer

To determine the value of $ k $, the y-coordinate of the point of intersection of the normals at points $ P $ and $ Q $ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we need to follow these steps:

1. **Equation of the Normal at a Point on the Hyperbola:**

For a point $ P(a \sec \theta, b \tan \theta) $ on the hyperbola, the equation of the normal is given by:
   $$
   ax \sec \theta + by \tan \theta = a^2 + b^2
   $$

Similarly, for the point $ Q(a \sec \phi, b \tan \phi) $, the equation of the normal is:
   $$
   ax \sec \phi + by \tan \phi = a^2 + b^2
   $$

2. **Condition $\theta + \phi = \frac{\pi}{2}$:**

Given that $\theta + \phi = \frac{\pi}{2}$, we have:
   $$
   \sec \phi = \csc \theta \quad \text{and} \quad \tan \phi = \cot \theta
   $$

3. **Substituting the Trigonometric Identities:**

Substitute $\sec \phi = \csc \theta$ and $\tan \phi = \cot \theta$ into the equation of the normal at $ Q $:
   $$
   ax \csc \theta + by \cot \theta = a^2 + b^2
   $$

4. **Solving the System of Equations:**

We now have two equations:
   $$
   ax \sec \theta + by \tan \theta = a^2 + b^2
   $$
   $$
   ax \csc \theta + by \cot \theta = a^2 + b^2
   $$

Solving these simultaneously for $ y $ (i.e., $ k $), we eliminate $ x $ by multiplying the first equation by $\csc \theta$ and the second by $\sec \theta$, and then subtracting:
   $$
   a(x \sec \theta \csc \theta) + by = (a^2 + b^2) \csc \theta
   $$
   $$
   a(x \csc \theta \sec \theta) + by = (a^2 + b^2) \sec \theta
   $$

Subtracting these equations gives:
   $$
   by (\tan \theta - \cot \theta) = (a^2 + b^2)(\csc \theta - \sec \theta)
   $$

Simplifying, since $\tan \theta - \cot \theta = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}$, we find:
   $$
   k = -\frac{a^2 + b^2}{b}
   $$

5. **Conclusion:**

Therefore, the correct value of $ k $ is $-\frac{a^2 + b^2}{b}$, which corresponds to Option 4.

**Why Other Options are Incorrect:**

- **Option 1 ($\frac{a^2 + b^2}{a}$)**: This option does not account for the negative sign and the division by $ b $, which is necessary due to the trigonometric identities and the subtraction process used to eliminate $ x $.

- **Option 2 ($\frac{a^2 + b^2}{b}$)**: This option misses the negative sign, which arises from the subtraction of the equations and the trigonometric identities involved.

- **Option 3 ($-\frac{a^2 + b^2}{a}$)**: This option incorrectly uses $ a $ in the denominator instead of $ b $, which does not align with the derivation process.

Thus, Option 4 is the correct answer, as it accurately reflects the derived value of $ k $.