Question 1080724
The slope of the tangent line is equal to the value of the derivative at that point.
{{{y^2/4-x^2/2=1}}}
{{{y^2-2x^2=4}}}
Implicitly differentiate,
{{{2ydy-4xdx=0}}}
{{{2ydy=4xdx}}}
{{{dy/dx=2(x/y)}}}
So then,
{{{y^2-2(4)^2=4}}}
{{{y^2-32=4}}}
{{{y^2=36}}}
{{{y=6}}} and {{{y=-6}}}
So for (4,6),
{{{m=dy/dx=2(4/6)=4/3}}}
Using the point slope form of a line,
{{{y-6=(4/3)(x-4)}}}
{{{y-6=(4/3)x-16/3}}
{{{y=(4/3)x-16/3+18/3}}}
{{{y=(4/3)x+2/3}}}
and for (4,-6)
{{{m=-4/3}}}
and
{{{y+6=-(4/3)(x-4)}}}
{{{y+6=-(4/3)x+16/3)}}}
{{{y=-(4/3)x+16/3-18/3)}}}
{{{y=-(4/3)x-2/3}}}
To get the normal lines, you know that the tangent and normal lines are perpendicular to each other.
So the slopes are negative reciprocals,
{{{m[1]*(4/3)=-1}}}
{{{m[1]=-3/4}}}
and
{{{m[2]*(-4/3)=-1}}}
{{{m[2]=3/4}}}
So then,
(4,6)
{{{y-6=(3/4)(x-4)}}}
(4,-6)
{{{y+6=(3/4)(x-4)}}}
I leave those to you to put into slope-intercept form.

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*[illustration 234.JPG].