Find the equations of the chords of the parabola y2 = 4ax which pass through the point (-6a, 0) and which subtends an angle of 45° at the vertex.

The vertex of parabola
y^2 = 4ax
is (0,0).
 Equation of a chord in the slope-intercept form is
y=kx+b (1)
Here k=tan 45°=1, because a chord subtends an angle of 45° at the vertex.
So, in fact (1) is given by
y=x+b (2)
On the other hand, this line passes through the point (–6a, 0), consequently its coordinates satisfy equation
(2):
0=-6a+b,
b=6a.
Finally, y=x+6a is the equation of chord.
Answer: y=x+6a.

The vertex of parabola y²=tax is O (0,0)

Equation of the chord is

y-y₁=m (x-x₁)

The chord passes through point  (-6a,0)

Thus point  should satisfy the eqⁿ  of chord 

y=m (x+6a)

Now ,y²=4ax (y-mx)/(6ma)-eq1

Since , (y-mx)/6ma=1

Now slove the equation 1you will get

m₁+m₂=2/3m 

m₁m₂=2/3

We know that, 

tan (A)=|m₁-m₂|/|1+m₁m₂|

Solve that by squaring and you will get ,

m=+2/7,-2/7

The equation of chord will be 1]y=2/7 (x+6a)

                                                     2]y=-2/7 (x+6a)