The slope of the line touching both the

parabolas y2 = 4x and x2 = 32y is

Lets take a general parabola : y² = 4ax
$$\begin{gathered} On differentiating it , we get \\ 2y\frac{dy}{dx}= 4a \\ Or \frac{dy}{dx} =\frac{2a}{y} \\ And at (x_1 ,y_1 ) , \frac{dy}{dx} =\frac{2a}{y_1} =m \\ So y_1 =\frac{2a}{m} and for y^2 =4ax \\ Then (\frac{2a}{m}{)}^2 =4a{x}_1, so x_1 =\frac{a}{m^2} \\ Hene the eqaution of \tan gent is (y-y_1) = m (x-x_1) \\ y -\frac{2a}{m} =m(x -\frac{a}{m^2}) \\ Or y =mx +\frac{a}{m} (equation of \tan gent for y^2 = 4ax ) \end{gathered}$$

$$\begin{gathered} Lets take x^2 = 4ay \\ Upon differentiating , we have \\ 2x =4a\frac{dy}{dx} \\ So \frac{dy}{dx} =\frac{x}{2a} \\ And at (x_2 , y_2 ) , \frac{dy}{dx} = \frac{x_2}{2a } = m \\ So x_2 = 2am \\ And (2am{)}^2 =4a{y}_2 \\ 4a^2{m}^2 =4a{y}_2 \\ Or y_2 =a{m}^2 \\ hence the eqaution of \tan gent is y-a{m}^2 = m(x-2am) \\ Or y =mx -a{m}^2 ( equation of \tan gent at x^2 = 4ay ) \end{gathered}$$

So our curve are  y² = 4x ( here a = 1) and x² = 32y ( here a = 8)
And the tangent to both curves are same
Hence  y = mx + 1/m and y = mx - 8m²
So on solving , we have m³ = -1/8
Hence m = -1/2