Option (C) is correct.

Given parabola is y² = 4x ....(1)

Let the point from where tangents are drawn be (h, k). i.e., (h, k) is point of intersection of two tangents to parabola y² = 4x.

Tangents of parabola (1) is T : y = mx + \(\frac{1}{m} .....(2)\)

(h, k) lies on T.

∴  k = mh + \(\frac{1}{m}\)

\(\Rightarrow m^2h - mh + 1 = 0 .....(3)\)

which is a quadratic equation in m.

Let m₁ and m₂ are two roots of quadratic equation (3)

∴ \(m_1 + m_2 = \frac{-(-k)}{h} = \frac{k}{h}\) and \(m_1m_2 = \frac{1}{h}.\)

∴ \((m_1 - m_2)^2\) \(= (m_1 + m_2)^2 - 4m_1m_2\)

\(= \left(\frac{k}{h}\right)^2 - \frac{4}{h}\)

\(= \frac{k^2 - 4h}{h^2}\)

∴ \(m_1 - m_2 = \frac{\sqrt{k^2 - 4h}}{h}.\)

Let angle between two tangent be α.

∴ tan α \(= \pm \frac{m_1 - m_2}{1 + m_1m_2}\) \(= \pm \cfrac{\frac{\sqrt{k^2 - 4h}}{h}}{1 + \cfrac{1}{h}}\)

\(\Rightarrow\) tan α \(= \pm \frac{\sqrt{k^2 - 4h}}{h + 1}\)

\(\Rightarrow\) k² - 4h = tan²α (h + 1)² .....(4)

But given that tangents include an angle of \(45^\circ.\)

i.e., α = \(45^\circ\)

\(\Rightarrow\) tan α = tan \(45^\circ\) = 1

∴ k² - 4h = (h + 1)² (From (4))

∴ Locus of point (h, k) is y² - 4x = (x + 1)² 

\(\Rightarrow\) y² - 4x = x² + 2x + 1

\(\Rightarrow\) x² - y² + 6x + 1 = 0.