Tangent of Locus

Let the coordinates of $A$ be $(x_A, 0)$ , and for $B$ $(x_B, 0)$ , and those of $P$ be $(x, y)$ , where $P$ is any point on the desired curve. Further, let $a = PB$ and $b = PA$ . We have

$a = \sqrt{ (x - x_B)^2 + y^2 } \text{ and} b = \sqrt{ (x - x_A)^2 + y^2 }$

The condition on the curve can be expressed as: $y + a + b = k$ is a constant.
Differentiating this condition implicitly with respect to x, we get

$y' + \dfrac{1}{a} ( (x - x_B) + y y' ) + \dfrac{1}{b} ( (x - x_A) + y y' ) ) = 0 \quad (1)$

Now when point $P$ coincides with point $C = (x_C, y_C)$ ( of $\triangle ABC$ ), we have the following from the given figure,

$(x_C - x_B) = a \cos 75^{\circ} , y_C = a \sin 75^{\circ} = b \sin 15^{\circ}, (x_C - x_A) = - b \cos 15^{\circ}$

Substituting the above three equations in equation (1) , we get at the point $C$ ,

$y'(1 + \sin 75^{\circ} + \sin 15^{\circ} ) + \cos 75^{\circ} - \cos 15^{\circ} = 0 \quad (2)$

Solving for y',

$y' = \dfrac{\cos 15^{\circ} - \cos 75^{\circ}} {1 + \sin 75^{\circ} + \sin 15^{\circ}}$

The reciprocal of that is

$\dfrac{1}{y'} =\dfrac{1 + \sin 75^{\circ} + \sin 15^{\circ}}{\cos 15^{\circ} - \cos 75^{\circ}}$

We now use the sum and difference trigonometric formulas :
$\cos A - \cos B = - 2 \sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$ and $\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A - B }{2}$

Hence, $\cos 15^{\circ} - \cos 75^{\circ} = - 2 \sin 45^{\circ} \sin (-30^{\circ}) = \dfrac{1}{\sqrt{2}}$ and $\sin 15^{\circ} + \sin 75^{\circ} = 2 \sin 45^{\circ} \cos 30^{\circ} = \dfrac{\sqrt{3}}{\sqrt{2}}$

Therefore, $\dfrac{1}{y'} = \dfrac{1 + \dfrac{\sqrt{3}}{\sqrt{2}} }{\dfrac{1}{\sqrt{2}} } = \sqrt{2} + \sqrt{3}$
The answer is $2 + 3 = 5.$