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If $ c > \frac {1}{2}, $ how many lines through the point (0, c) are normal lines to the parabola $ y = x^2? $ What if $ c \le \frac {1}{2}? $

we have three normal lines if $c>\frac{1}{2}$ and one normal line if $c \leq \frac{1}{2}$

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I can't square. So when you right here. So we have Why is equal to X square? This is our given function. We're gonna differentiate it to get this slope off the tangent and we have two X We have the cornet zero comma SSI just given to us. The normal line is the line perpendicular to the tangent line. So we have to find a point or the tangent line passes. So we make X equal to a When we get the point A comma a square on the slope of the tangent line is too late. We have Where next? Gonna find the normal line, which is the perpendicular line on the slope is the negative reciprocal of the tension Learn slope So we get them off. The normal line is equal to negative one over, Um and this is equal to negative one over two A. So, after we find are a normal mind slow. We have the cornet points, that pastor, the tangent line and the normal line So we can now plug it in. We get why too minus y one over x two minus x one is equal to negative one over two a we plug it in a square. Runa C. Over a minute zero is equal to a negative one over two A. You'll get a square is equal to C minus 1/2. So if C is equal to one, have that means A is going to be equal to zero. So we get a square is equal to 1/2 minus one have is equal to zero. So there's only one solution. If see, it's bigger than zero and there are a can be positive or negative square root of seeing minus 1/2. So there are two solutions. And if C is less than 1/2 there is no solution. We're gonna drop her if we look at our parabola here. CIA's bigger, then 1/2. There are three normal lines, including the lie axis, and when C is less than or equal to negative 1/2 there's only one solution, and that's when C is equal to 1/2