specificagent1
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No clue how to do question b
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No clue how to do question b
saltshaker
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Is it not the same as (a), but using a different set of points? Assuming you already know a
specificagent1
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no because it's from an external point so there can be 2 possible tangents
You should probably have a look at the link below. I believe it may help:
Let gradient be 
First we will sub the point (1,-2) and into point gradient formula
=m(x-1)\\y=mx-(m+2))
Since the line and the parabola touch each other, we have to solve them simultaneously
\longrightarrow \textcircled{1}\\y=x^2\longrightarrow \textcircled{2})
Sub
into 
\\x^2-mx+(m+2)=0)
Using quadratic formula we can solve for x and
}}{2}\\x=\frac{m\pm\sqrt{m^2-4m-8}}{2})
However since the original line is tangent to the curve, there would only be one value for x, hence
which means that 
}}{2}\\\\m=\frac{4\pm\sqrt{48}}{2}\\\\m=2\pm2\sqrt{3})
Therefore the two equations for the line are...
(x-1)-2\ \text{and}\ y=(2+2\sqrt{3})(x-1)-2)
First we will sub the point (1,-2) and
Since the line and the parabola touch each other, we have to solve them simultaneously
Sub
Using quadratic formula we can solve for x and
However since the original line is tangent to the curve, there would only be one value for x, hence
Therefore the two equations for the line are...
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