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.

Question

Arun · Answer

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.

Harsh · Answer

The vertex of parabola y 2 =tax is O (0,0) Equation of the chord is y-y 1 =m (x-x 1 ) The chord passes through point (-6a,0) Thus point should satisfy the eq n of chord y=m (x+6a) Now ,y 2 =4ax (y-mx)/(6ma)-eq1 Since , (y-mx)/6ma=1 Now slove the equation 1you will get m 1 +m 2 =2/3m m 1 m 2 =2/3 We know that, tan (A)=|m 1 -m 2 |/|1+m 1 m 2 | 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)

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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.

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.

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