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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.

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3 years ago

```							The vertex of parabolay^2 = 4axis (0,0). Equation of a chord in the slope-intercept form isy=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 byy=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.
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3 years ago
```							The vertex of parabola y2=tax is O (0,0)Equation of the chord isy-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)-eq1Since , (y-mx)/6ma=1Now slove the equation 1you will getm1+m2=2/3m m1m2=2/3We know that, tan (A)=|m1-m2|/|1+m1m2|Solve that by squaring and you will get ,m=+2/7,-2/7The equation of chord will be 1]y=2/7 (x+6a)                                                     2]y=-2/7 (x+6a)
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one year ago
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