**Solution**

**The usual slope form gives only part of the picture.**  
Since m must be finite, it misses x = 0. Parametrization captures all common tangents — including the hidden vertical one.

For the standard parabola y² = 4ax,

the parametric point and tangent corresponding to parameter t are 

P = (at², 2at) and ty = x + at².

Here a = 1. Therefore, P = (t², 2t) and the tangent is x − ty + t² = 0.

The circle is (x − 4)² + y² = 16, with centre C = (4, 0) and radius 4.

For the tangent to the parabola to also be tangent to the circle,

Distance of C from the tangent = 4

⇒ |4 + t²| / √(1 + t²) = 4

Squaring,

(4 + t²)² = 16(1 + t²)

⇒ t⁴ + 8t² + 16 = 16 + 16t²

⇒ t⁴ − 8t² = 0

⇒ t²(t² − 8) = 0

Hence,

t = 0, ±2√2.

So the corresponding points of contact on the parabola are

P₁ = (0, 0)

P₂ = (8, 4√2)

P₃ = (8, −4√2).

Thus, the common tangents touch the parabola at three distinct points, and these points form a triangle.

Take P₂P₃ as the base.

P₂P₃ = 8√2.

Since P₂ and P₃ lie on x = 8, the perpendicular distance of P₁ = (0, 0) from P₂P₃ is 8.

Therefore,

Area = 1/2 × 8√2 × 8

= 32√2.

**Answer: 32√2**





Slope Form vs Parametric Form | The Hidden Vertical Tangent