Let foci be  (x₁, y₁)  and  (x₂, y₂)  and centre be (h, k)

$\therefore$   2h = x₁ + x₂ and 2k = y₁ + y₂

        here      x₁x₂ = y₁y = b₂

 $\because$  SS’ = 2ae

 ⇒    (x₁ – x₂)² + (y₁ – y₂)² = 4a²e²

 ⇒    (x₁ + x₂)² + (y₁ – y₂)² – 4x₁x₂ – 4y₁y₂ = 4(a₂ – b₂)

 ⇒    4h² + 4k² = 4(a² + b²)

 ⇒    h₂ + k₂ = ${\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$\therefore$   Locus is x² + y² = 1                                i.e. circle.

        Normal to ellipse $\frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{4}=1$ will be 3x cos θ – 2y cos θ = 5                                ....(i)

$\because$   (i) is tangent to director circle.

$\therefore$   Perpendicular distance from centre

          = radius of director circle

⇒        $\frac{5}{\sqrt{9{\sec }^2\mathrm{\theta }+4{\mathrm{cosec}}^2\mathrm{\theta }}}=\sqrt{2}$ 

$\because$   9 sec² θ + 4 cosec² θ

        = 13 + 9 tan² θ + $\frac{4}{{\tan }^2\mathrm{\theta }}\ge 25$

$\therefore$   maximum value of LHS$=1\ne \sqrt{2}$

$\therefore$   no such θ exists.

$\therefore$   number of tangents = 0 Ans.