The equation for you derivative is incorrect. Using implicit differentiation I get:

goes to

Solve for .

I get that the slopes of the line for x = 1 are and .

For the -1 slope, the slope of the normal line is +1. I get the line y = x. (See the graph below. This point and line is green.)

For the 0 slope, the slope of the normal is undefined. I get the line x = 1. (This is the blue point and line. Never mind that the blue line also intersects the green point. That's just an artifact of this normal line being parallel to the y-axis.

Sure you do! It's just that you have to take things step by step.

You know the equation of the normal line, thus you know the slope of the normal line. (Let's just consider the y= x normal line for simplicity.) So you are looking for a point on the ellipse where the normal line has a slope of 1. Thus the tangent line has a slope of -1. Thus:

Solve this equation for y and you have the equation of the tangent line that intersects the ellipse where the normal line has a slope of -1. I get the line y = x.

Now you want to find the points where this line intersects the ellipse. So plug y = x into your ellipse formula and solve for x. (There will be typically be two x values also, one of which you already know.) Then you can find the y value for these points by plugging x into y = x. Generally the two points willnotbe on the same line, as they are in this case.

This is just the same problem as b) but for the (1, -2) point. Do the same procedure.

-Dan