Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
I. x2 + xy - y2 = 1, (2,3) (10)
II x2 + y2 = 25, (3,-4)
III. x3y2 = 9, (-1,3)
IV. y2 – 2x – 4y - 1 = 0, (-2, 1)

For each curve, plug in the given point and check if the equality holds. For example:

(I) (2, 3) does lie on since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative , and evaluate it for the given point . This is the slope of the tangent line at the point. For example:

(I) The derivative is

so the slope of the tangent at (2, 3) is

and its equation is then

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, . For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.

i think the answer is b