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.

Arun
25750 Points
6 years ago
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.
Harsh
15 Points
4 years ago
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 eq 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)