TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the given parabola.
y2=4x
The general equation of parabola is given by,
y=mx+‌

a

m

At point (1,4) the equation becomes,
4‌‌=m+‌

1

m

m2−4m+1‌‌=0
Let the roots of the equationbe m1 and m2
m1+m2=4
And
m1m2=1
Therefore
tan‌θ‌‌=‌

|m1−m2|

|1+m1m2|

‌‌=‌

√(m1+m2)2−4m1m2

1+m1m2

‌‌=‌

√42−4

1+1

‌‌=‌

√12

2

Solve further
tan‌θ‌‌=‌

2√3

2

‌‌=√3
θ‌‌=tan−1(√3)
‌‌=‌

π

3