Here, the given equation of parabola is y2= 8x.
The equation of tangent to the parabola y2=8x is,
y= mx + 2/m
This tangent passes through the point (-2, 3)
So, 3 = -2m + 2/m
or, 3m + 2m2 = 2
or, 2m2+3m - 2= 0
or, 2m2 + (4 - 1)m -2 = 0
or, 2m2 + 4m - m - 2 = 0
or, 2m(m + 2) - 1(m+2) = 0
or, (m + 2) (2m - 1) = 0
Either, Or,
m = -2 m = 1/2
Required angle is,
Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
| n | a(-ve) | b(+ve) | cn | f(cn) |
| 0 | 2 | 3 | 2.5 | 0.8623 |
| 1 | 2 | 2.5 | 2.25 | -0.050595 |
| 2 | 2.25 | 2.5 | 2.375 | 0.38664 |
| 3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
| 4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
| 5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
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