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Find the equation of the normal to the parabola `y^2 = 4x` which is parallel to `y - 2x + 5 = 0`A. y=2x+12B. y=2x-12C. y=2x+8D. y=-2x+12 |
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Answer» Correct Answer - B The equation of the normal to the parabola `y^(2)=4ax" at "(am^(2), -2m)` is `y=mx-2am-am^(3)`, where m is the slope of the normal. Here, a = 1. So, the equation of the normal at `(m^(2), -2m)` is `y = mx-2m -m^(3)" ....(i)"` If the normal is parallel to the line y=2x-5. Then, m = Slope of the line `y = 2x - 5 rArr m=2` Putting the value of m in (i), we obtain y=2x-12 as the equation of the requaired normal. |
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