Calculus (1 Viewer)

[hs17]

a) Find the x-coordinates of the points P and Q on y = (x − 7)^2 +3 such that the tangents at P and Q have gradients 1 and −1 respectively.

b) Show that the square formed by the tangents and normals at P and Q has area 1/2

I got part a ...

The coordinates are P= (7.5, 3.25), Q= (6.5, 3.25)

But I have no idea how to do part b

.

 

[TheShy]

For questions like this, its best to actually graph if it since it becomes easier once you visually know whats going on. To do part b, you need to find the equation of the tangents P and Q. So just sub the x and y values into y -y1 = m(x-x1) and you would get the equations. Then since the question asks for the square formed by both the tangent and normals at P and Q, you would need to find the tangents. Again using y-y1 = m(x-x1). Then use distance formula to find the size of one of the sides. Once you have that just square that side to find the area.

However, an easier way is to find the intersection point of the tangents of P and Q. Once you have that, use distance formula for the instersection point and a coordinate P or Q. If you do that, you would get the size of one of the sides of the square. Square that to get the area.

 

[hs17]

OMG thank you!!! i hope i'm not bothering you too much with all these questions haha

 

[TheShy]

Lmao its g dw

 

[CM_Tutor]

Given the coordinates
and
you know that the distance and you know that must be the diagonal of the square (it can't be a side as is horizontal but the sides of the square have gradients of and ).

Now, for any square of side length with diagonal length , Pythagoras' Theorem gives us that , and since the area of a square is , we can conclude that:



and since we know here that , it follows that:



If you were not told that the two normals and tangents enclosed a square, you could deduce it geometrically as follows:

• The angle between the tangent at and the normal at is a right angle from the definition of a normal at a point on a curve as the line perpendicular to the tangent at that point.
• It follows that the angle between the tangent at and the normal at is also a right angle
• We know that the tangents at and have gradients of and , respectively, and since , these two tangents are perpendicular
• So, our quadrilateral is either a square or a rectangle (as all four of its angles are 90 degrees)
• The midpoint of is at , and it must also be the midpoint of the other diagonal, , whether the quadrilateral is a square or a rectangle
• If the shape is a square, the diagonals must be perpendicular, and since is horizontal, that would make vertical, so let's try for and to lie on
• Now, the tangent at has a gradient of 1 and so any change in will be accompanied by the same change in - that is
• Starting from , we need to move to , making , and putting at
• does lie on the tangent at (evident since that tangent has gradient -1 and so a requires , making the shift from to : as required
• We immediately know the coordinates of are because is the midpoint of , but this could also be established by a similar method using the two normals
• We have confirmed that is a square as (as is horizontal and is vertical)
• We have the area as half the square of the diagonal, or we could show that all four sides have the same length and then say that the area is