Let P($$3\sec\theta, 2\tan\theta$$) and Q($$3\sec\phi, 2\tan\phi$$) where $$\theta + \phi = \frac{\pi}{2}$$, be two distinct points on the hyperbola $$\frac{x^2}{9} - \frac{y^2}{4} = 1$$. Then the ordinate of the point of intersection of the normals at P and Q is:

We have the rectangular hyperbola in standard form

$$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$

whose semi-transverse axis is $$a=3$$ and semi-conjugate axis is $$b=2$$. For this curve the standard parametric equations are

$$x=a\sec\alpha,\qquad y=b\tan\alpha.$$

Hence the points

$$P\;(3\sec\theta,\,2\tan\theta),\qquad Q\;(3\sec\phi,\,2\tan\phi)$$

indeed lie on the hyperbola. It is given that

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

To find the intersection of the normals at these two points, we first write the slope of the normal at a general point.

The hyperbola is represented by $$F(x,y)=\frac{x^{2}}{9}-\frac{y^{2}}{4}-1=0.$$ Differentiating implicitly with respect to $$x$$,

$$\frac{2x}{9}-\frac{2y}{4}\,\frac{dy}{dx}=0 \;\;\Longrightarrow\;\;\frac{dy}{dx}=\frac{4x}{9y}.$$

Thus the slope of the tangent at $$(x,y)$$ is $$m_t=\dfrac{4x}{9y}$$, and the slope of the normal is the negative reciprocal:

$$m_n=-\frac{1}{m_t}=-\frac{9y}{4x}.$$

Applying this to the two parametric points:

For $$P(3\sec\theta,2\tan\theta):$$

$$m_P=-\frac{9(2\tan\theta)}{4(3\sec\theta)} =-\frac{18\tan\theta}{12\sec\theta} =-\frac{3\tan\theta}{2\sec\theta} =-\frac{3\sin\theta}{2}.$$

For $$Q(3\sec\phi,2\tan\phi):$$

$$m_Q=-\frac{9(2\tan\phi)}{4(3\sec\phi)} =-\frac{3\tan\phi}{2\sec\phi} =-\frac{3\sin\phi}{2}.$$

Now we write the equations of the two normals.

Through $$P(3\sec\theta,2\tan\theta):$$

$$y-2\tan\theta=m_P\bigl(x-3\sec\theta\bigr).$$

Through $$Q(3\sec\phi,2\tan\phi):$$

$$y-2\tan\phi=m_Q\bigl(x-3\sec\phi\bigr).$$

Re-arranging each into the slope-intercept form $$y=mx+c$$:

For $$P$$ (putting $$m_1=m_P$$): $$$ y=m_1x+\underbrace{\bigl(2\tan\theta-3m_1\sec\theta\bigr)}_{c_1}. $$$

For $$Q$$ (putting $$m_2=m_Q$$): $$$ y=m_2x+\underbrace{\bigl(2\tan\phi-3m_2\sec\phi\bigr)}_{c_2}. $$$

Let us evaluate the two intercepts explicitly.

Because $$m_1=-\dfrac{3\sin\theta}{2}$$,

$$$ c_1=2\tan\theta-3m_1\sec\theta =2\frac{\sin\theta}{\cos\theta} -3\Bigl(-\frac{3\sin\theta}{2}\Bigr)\frac{1}{\cos\theta} =\frac{4\sin\theta}{\cos\theta}+\frac{9\sin\theta}{2\cos\theta} =\frac{13\sin\theta}{2\cos\theta} =\frac{13}{2}\tan\theta. $$$

Similarly, with $$m_2=-\dfrac{3\sin\phi}{2}$$,

$$$ c_2=2\tan\phi-3m_2\sec\phi =\frac{13}{2}\tan\phi. $$$

The intersection of the two straight lines

$$y=m_1x+c_1,\qquad y=m_2x+c_2$$

has ordinates obtained from the determinant formula

$$Y=\frac{m_1c_2-m_2c_1}{m_1-m_2}.$$

Substituting the explicit expressions,

$$$ Y=\frac{\displaystyle m_1\Bigl(\frac{13}{2}\tan\phi\Bigr) -m_2\Bigl(\frac{13}{2}\tan\theta\Bigr)} {m_1-m_2} =\frac{13}{2}\; \frac{m_1\tan\phi-m_2\tan\theta}{m_1-m_2}. $$$

Insert the values $$m_1=-\dfrac{3\sin\theta}{2},\; m_2=-\dfrac{3\sin\phi}{2}$$:

$$$ m_1\tan\phi=-\frac{3\sin\theta}{2}\,\frac{\sin\phi}{\cos\phi} =-\frac{3\sin\theta\sin\phi}{2\cos\phi}, $$$

$$$ m_2\tan\theta=-\frac{3\sin\phi}{2}\,\frac{\sin\theta}{\cos\theta} =-\frac{3\sin\phi\sin\theta}{2\cos\theta}. $$$

Therefore

$$$ m_1\tan\phi-m_2\tan\theta =-\frac{3\sin\theta\sin\phi}{2}\Bigl(\frac{1}{\cos\phi}-\frac{1}{\cos\theta}\Bigr). $$$

And

$$$ m_1-m_2=-\frac{3\sin\theta}{2}+\frac{3\sin\phi}{2} =\frac{3}{2}\bigl(\sin\phi-\sin\theta\bigr). $$$

Substituting back,

$$$ Y=\frac{13}{2}\; \frac{\displaystyle -\dfrac{3\sin\theta\sin\phi}{2} \Bigl(\dfrac{1}{\cos\phi}-\dfrac{1}{\cos\theta}\Bigr)} {\displaystyle \dfrac{3}{2}\,(\sin\phi-\sin\theta)} =-\frac{13}{2}\, \frac{\sin\theta\sin\phi\;\bigl(\frac{1}{\cos\phi}-\frac{1}{\cos\theta}\bigr)} {\cos\theta\cos\phi\;(\, \frac{\sin\phi-\sin\theta}{\cos\theta\cos\phi}\,)} =-\frac{13}{2}\, \frac{\sin\theta\sin\phi\;(\cos\theta-\cos\phi)} {\cos\theta\cos\phi\;(\sin\phi-\sin\theta)}. $$$

At this point we employ the crucial relation $$\theta+\phi=\dfrac{\pi}{2}$$, which implies

$$\sin\phi=\cos\theta,\qquad \cos\phi=\sin\theta.$$

Putting these into the last expression, every factor cancels:

$$$ Y=-\frac{13}{2}\, \frac{(\sin\theta)(\cos\theta)\;(\cos\theta-\sin\theta)} {(\cos\theta)(\sin\theta)\;(\cos\theta-\sin\theta)} =-\frac{13}{2}. $$$

Thus the ordinate (the $$y$$-coordinate) of the point where the two normals meet is

$$Y=-\frac{13}{2}.$$

Hence, the correct answer is Option D.