(i)
y2 = 8x
4a = 8
⇒ a = 2
\(\therefore\) Focus is (a, 0) i.e., (2, 4) & (2, -4).
And End points of latus rectum are
(a, 2a) & (a, -2a) i.e., (2, 4) & (2, -4).
\(\therefore\) Lower end point of latus rectum is (2, -4).
\(\therefore\) Required line is line joining point (0, 0) & (2, -4).
\(\therefore\) y = mx be such line as it passes through origin.
\(\therefore\) -4 = 2m (\(\because \) (2, -4) satisfies equation of line y = mx)
\(\therefore\) m = -2
\(\therefore\) Equation of required line is y = -2x.
(ii)
Let point on parabola is (x1, y1).
\(\therefore {y_1}^2 = 4ax_1\)
\(\therefore x_1 = \frac{{y_1}^2}{4a}\)
\(\therefore \) Consider point is \(P(\frac{{y_1}^2}{4a}, y_1)\)
such that OP makes \(\theta \) angle with x-axis.
\(\therefore \) Slope of tangent of line OP = \(\frac{y_1 - 0}{\frac{{y_1}^2}{4a} -0} = \frac{y_1 }{\frac{{y_1}^2}{4a} } = \frac{4a}{y_1}\)
\(\because \) OP makes \(\theta \) angle with x-axis.
\(\therefore \) Slope of tangent = \(\tan\theta \)
\(\therefore \frac{4a}{y_1} = \tan\theta\)
\(\therefore y_1 = \frac{4a}{\tan \theta} = 4a\cot \theta \)
\(x_1 = \frac{{y_1}^2}{4a} = \frac{16a^2 \cot^2\theta}{4a} = 4a\cot ^2\theta\)
\(\therefore \) Length of line segment \(OP^2 = {x_1}^2 + {y_1}^2\)
\(= 16a^2 \cot^4\theta + 16a^2\cot^2\theta\)
\(= 16a^2\cot^2\theta (\cot^2\theta + 1)\)
\(= 16a^2 \cot^2\theta \,cosec^2\theta\)
\(\therefore OP = 4a \cot \theta.cosec\theta\)
