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[U MAD BRO]

This is a 3U question but I found it in a 4U paper in question 5, it is worth 3 marks.
I tried to do it, but looks like I'm not doing it right
Question: A thin wire of length A is divided into 2 pieces, from which a circle and a square are made. Show that the smallest possible area enclosed by the circle and the square is
can someone at least give me a hint please?

 

[Nooblet94]

Cut the wire into two pieces of length x and A-x respectively and make one into a circle, one into a square. Find the areas of the circle/square in terms of x and then differentiate to find the x-value at which the minimum occurs. Sub that back in and you'll get the answer.

 

[U MAD BRO]

thank you
one other question.
an ellipse with chord PQ, tangents at P and Q intersect at an external point T. Show that PQ is a focal chord if T lies on the directrix. Hence find the area of the triangle PQT if T also lies on the x-axis.
The working out for this is getting really messy and looks like im going no where

 

[Carrotsticks]

1) Find the equation of the chord PQ and sub in (ae,0). By doing this, you will find the condition such that PQ is a focal chord.

2) Find the coordinates of T in terms of P and Q.

3) Equate the x coordinate of T with the x coordinate of the directrix (so T lies on the directrix)

4) Acquire the same condition from (1)

Now for the area part:

If T lies on the x axis, then the two tangents will share the same x coordinate. The y coordinates will be the same in magnitude, but opposite in length.

The Base of the triangle is the length PQ (twice the y coordinate) and the height of the triangle is the x coordinate of T minus the x coordinate of PQ (draw a diagram, you will see).

Then proceed to use 1/2 base x height etc.

Alternatively, you could construct the 3x3 matrix (or 2x2 matrix by translation of one of the points to the origin), then find half the determinant =)

 

[U MAD BRO]

Thank you a lot Carrot =)