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 the ordinate of the point of intersection of normals at $$P$$ and $$Q$$ is $$-k\left(\frac{a^2+b^2}{2b}\right)$$, then $$k$$ is equal to _________.

The hyperbola is $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ and the two points on it are

$$$P\bigl(a\sec\theta,\;b\tan\theta\bigr),\qquad Q\bigl(a\sec\phi,\;b\tan\phi\bigr),\qquad\text{with }\;\theta+\phi=\dfrac{\pi}{2}.$$$

For the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ the slope of the tangent at a general point $$(x_{1},y_{1})$$ is obtained by implicit differentiation:

$$$\frac{2x_{1}}{a^{2}}-\frac{2y_{1}}{b^{2}}\frac{dy}{dx}=0 \;\Longrightarrow\; \frac{dy}{dx}=\frac{b^{2}x_{1}}{a^{2}y_{1}}.$$$

Hence the slope of the normal is the negative reciprocal:

$$m_{n}=-\frac{a^{2}y_{1}}{b^{2}x_{1}}.$$

Normal at $$P$$

For $$P(a\sec\theta,\;b\tan\theta)$$ we insert $$x_{1}=a\sec\theta,\;y_{1}=b\tan\theta$$ to get

$$$m_{1}=-\frac{a^{2}(b\tan\theta)}{b^{2}(a\sec\theta)} =-\frac{ab\tan\theta}{b^{2}\sec\theta} =-\frac{a}{b}\,\frac{\tan\theta}{\sec\theta} =-\frac{a}{b}\sin\theta.$$$

Thus the normal at $$P$$ is

$$$y-b\tan\theta =-\frac{a}{b}\sin\theta\;\bigl(x-a\sec\theta\bigr). \quad -(1)$$$

Normal at $$Q$$

Because $$\phi=\dfrac{\pi}{2}-\theta,$$ we have $$\sin\phi=\cos\theta$$, $$\; \tan\phi=\cot\theta$$, $$\; \sec\phi=\csc\theta.$$ The slope of the normal at $$Q$$ becomes

$$$m_{2} =-\frac{a^{2}(b\tan\phi)}{b^{2}(a\sec\phi)} =-\frac{a}{b}\frac{\tan\phi}{\sec\phi} =-\frac{a}{b}\sin\phi =-\frac{a}{b}\cos\theta.$$$

Therefore the normal at $$Q$$ is

$$$y-b\tan\phi =-\frac{a}{b}\cos\theta\;\bigl(x-a\sec\phi\bigr). \quad -(2)$$$

Point of intersection of the two normals

Let the intersection be $$(X,Y).$$ Using (1):

$$$Y=b\tan\theta-\frac{a}{b}\sin\theta\,(X-a\sec\theta). \quad -(3)$$$

Using (2) and substituting the trigonometric equivalents:

$$$Y=b\cot\theta-\frac{a}{b}\cos\theta\, \bigl(X-a\csc\theta\bigr). \quad -(4)$$$

Equating the right-hand sides of (3) and (4):

$$$b\tan\theta-\frac{a}{b}\sin\theta\,(X-a\sec\theta) =b\cot\theta-\frac{a}{b}\cos\theta\, \bigl(X-a\csc\theta\bigr).$$$

We first isolate $$X$$. Write the equation as

$$$\left(-\frac{a}{b}\sin\theta+\frac{a}{b}\cos\theta\right)X =b\cot\theta-b\tan\theta +\frac{a^{2}}{b}\sin\theta\sec\theta -\frac{a^{2}}{b}\cos\theta\csc\theta.$$$

Simplifying step by step:

Slope difference (denominator):

$$$m_{1}-m_{2} =-\frac{a}{b}\sin\theta-\Bigl(-\frac{a}{b}\cos\theta\Bigr) =-\frac{a}{b}\bigl(\sin\theta-\cos\theta\bigr).$$$

Numerator:

$$$b\bigl(\cot\theta-\tan\theta\bigr) +\frac{a^{2}}{b}\bigl(\cot\theta-\tan\theta\bigr) =\bigl(\cot\theta-\tan\theta\bigr)\frac{a^{2}+b^{2}}{b}.$$$

Hence

$$$X =\frac{\bigl(\cot\theta-\tan\theta\bigr)\dfrac{a^{2}+b^{2}}{b}} {-\dfrac{a}{b}\bigl(\sin\theta-\cos\theta\bigr)} =-\frac{a^{2}+b^{2}}{a}\; \frac{\cot\theta-\tan\theta}{\sin\theta-\cos\theta}.$$$ Call

$$$R=\frac{\cot\theta-\tan\theta}{\sin\theta-\cos\theta},\qquad \text{so that}\qquad X=-\frac{a^{2}+b^{2}}{a}\,R. \quad -(5)$$$

Value of $$R$$

$$$\cot\theta-\tan\theta =\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta} =\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta} =\frac{\cos2\theta}{\sin\theta\cos\theta}$$$

and

$$\sin\theta-\cos\theta =-(\cos\theta-\sin\theta).$$

Hence

$$$R =\frac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)} {-(\sin\theta\cos\theta)(\cos\theta-\sin\theta)} =-\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}. \quad -(6)$$$

Now the ordinate $$Y$$

From (3):

$$$Y =b\tan\theta-\frac{a}{b}\sin\theta\Bigl(X-a\sec\theta\Bigr).$$$

Using (5):

$$$-\frac{a}{b}\sin\theta\,X =-\frac{a}{b}\sin\theta\Bigl(-\frac{a^{2}+b^{2}}{a}R\Bigr) =\frac{a^{2}+b^{2}}{b}\sin\theta\,R.$$$

Also

$$$-\frac{a}{b}\sin\theta\bigl(-a\sec\theta\bigr) =\frac{a^{2}}{b}\sin\theta\sec\theta =\frac{a^{2}}{b}\tan\theta.$$$

Therefore

$$$Y =b\tan\theta+\frac{a^{2}+b^{2}}{b}\sin\theta\,R+\frac{a^{2}}{b}\tan\theta =\frac{a^{2}+b^{2}}{b}\tan\theta+\frac{a^{2}+b^{2}}{b}\sin\theta\,R.$$$

Factor $$\dfrac{a^{2}+b^{2}}{b}$$:

$$$Y=\frac{a^{2}+b^{2}}{b}\Bigl[\tan\theta+\sin\theta\,R\Bigr]. \quad -(7)$$$

The bracketed constant

Using $$R$$ from (6):

$$$\sin\theta\,R =\sin\theta\Bigl[-\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\Bigr] =-\frac{\cos\theta+\sin\theta}{\cos\theta}.$$$

Hence

$$$\tan\theta+\sin\theta\,R =\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta+\sin\theta}{\cos\theta} =\frac{\sin\theta-\cos\theta-\sin\theta}{\cos\theta} =-\frac{\cos\theta}{\cos\theta} =-1.$$$

Thus the ordinate simplifies to

$$$Y =\frac{a^{2}+b^{2}}{b}\,(-1) =-\frac{a^{2}+b^{2}}{b} =-2\left(\frac{a^{2}+b^{2}}{2b}\right).$$$

Comparing with the given form $$Y=-k\left(\dfrac{a^{2}+b^{2}}{2b}\right)$$ we immediately read

$$k=2.$$

So, the answer is $$2$$.