A tangent to the parabola y² = 4x is inclined at an angle 45° deg with the positive direction of x-axis. What is the point of contact of the tangent and the parabola?

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

\(m \;=\; \tan(45^\circ) \;=\; 1.\)

A standard parametric form for y2 = 4x is 

\(\bigl(x(t),\,y(t)\bigr) \;=\;\bigl(t^{2},\,2t\bigr), \)

since  \(y^{2} = 4x \implies (2t)^{2} = 4\,t^{2} \)

The slope of the tangent at \(\bigl(t^{2},\,2t\bigr) \)

Differentiate y2 = 4x 

\(2y\,\frac{dy}{dx} \;=\; 4 \)

\(\;\Longrightarrow\; \frac{dy}{dx} \;=\; \frac{4}{\,2y\,} \;=\; \frac{2}{\,y\,}. \)

At the point \(\bigl(x,y\bigr) = \bigl(t^{2},\,2t\bigr) \) one has \(y = 2t \)

\(\left.\frac{dy}{dx}\right|_{\,(t^{2},\,2t)} \) \(= \frac{2}{\,2t\,} = \frac{1}{t}.\)

We require this slope to equal 1.

\(\frac{1}{t} \;=\; 1 \;\Longrightarrow\; t = 1. \)

Now point of contact

Substitute t = 1 in \(\bigl(x(t),\,y(t)\bigr) = \bigl(t^{2},\,2t\bigr)\). 

\(x(1) = 1^{2} = 1, \)

\(y(1) = 2 \cdot 1 = 2. \)

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.