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The equation of a tangent to the para...

The equation of a tangent to the parabola `y^2=""8x""i s""y""=""x""+""2` . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is (1) `(-1,""1)` (2) `(0,""2)` (3) `(2,""4)` (4) `(-2,""0)`

Text Solution

Verified by Experts

Given, `lambda=5mm=5xx10^-3m`
` k=(2pi)/lambda =(2pi)/(5xx10^-3)m^-1`
` = 1257m^-1`
From the equation,
`c = omega/k`
`omega = c k= (3xx10^8)(1257)`
`3.77xx10^11 rad//s`
`E_0=30 V//m`
`c = E_0/B_0`
`B_0=E_0/c = 30/3xx10^8`
`10^-7 T`
Now,
`E_y=E_0 sin (omegat-kx)`
`(30 V// m) sin [3.77xx10^11 s^-1)t-(1257m^-1)x]`
and ` B_z=(10^-7 T) sin [3.77xx10^11 8^-1)t-(1257,^-1)x]`.
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