Tangent ProblemIts givin me a really really bad headache

[saltrock]

the curve with the equation y=e^x/2 meets the y-axis at point A

a)Find the equation of tangent
This is easy as pie..I have got the equation as 2y=x+1

b)The point B has the x co-ordinate ln4 and lies on the curve.The normal at B to the curve meets the tangent at A to the curve at the point C.
provr that x co ordinate of C is 3/2+ ln 2 and find the y co ordinate of C
>>>>This question is really killing me ...Any help would really be appriciated

c)Find in terms of e,the area of the finite region bounded by the curve with the equation y=1/2 e^x,the co ordinate axis and the line with the equation x+2=0
>>>>this looks easy..just gimme some hint please

 

[daster]

b) Find the (x, y) coordinates of A and B (at A x=0 and at B x=ln4). Find the equation of the normal to the curve at B (the slope of the normal is (1/y'), where y is the equation of the curve). The normal at B and the tangent at A intersect at point C; what does that tell you about C?

c) Draw a sketch of y=0.5e^x and x=-2. Where does the line x=-2 intersect the curve? Where does the curve intersect the y-axis? (The answers to these are the bounds of the integral that gives you the area.)

 

[M98Ranger]

a) To find the equation of the tangent at point A, we first need to find the slope of the curve at that point. To do this, we can take the derivative of the curve with respect to x: y'= (1/2)e^x/2. Then, we can substitute the x-coordinate of point A (which is 0, since it lies on the y-axis) into the derivative to find the slope at that point: y' = (1/2)e^0 = 1/2. Now, we can use the point-slope form of a line to find the equation of the tangent: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the curve. In this case, we have y - 0 = (1/2)(x - 0), which simplifies to y = (1/2)x. However, the equation given in the problem (2y = x + 1) is not the same as this equation, so there may have been a mistake in the calculations.

b) To prove that the x-coordinate of point C is 3/2 + ln2, we can use the fact that the product of the slopes of two perpendicular lines is -1. We already know the slope of the tangent at point A (1/2), and we can find the slope of the normal at point B by taking the negative reciprocal of the derivative at point B. The derivative of the curve is y' = (1/2)e^x/2, so at point B (ln4), the slope of the normal is -2. This means that the equation of the normal at point B is y - y1 = -2(x - x1), where (x1, y1) is the point (ln4, e^ln4/2) = (ln4, 2). Now, we can find the point of intersection between the tangent at A (y = (1/2)x) and the normal at B (y = -2x + 2) by solving the system of equations. This gives us the point (3/2 + ln2, 1/2 + 2ln2). Therefore, the x-coordinate of point C is 3/2 + ln2. To find the y-coordinate, we can substitute this value of x into the equation