0

If $ax^2+2hxy+by^2=0$ be the two sides of a parallelogram and $px+qy=1$ is its one diagonal, prove that the other diagonal is $y(bp-hq)=x(aq-hp)$.

My Approach.

Here, $ax^2+2hxy+by^2=0$ represents aa pair of lines passing through the origin. $$ax^2+2hxy+by^2=0$$ Multiplying both sides by $a$ $$a^2x^2+2ahxy+aby^2=0$$ $$a^2x^2+2ahxy+h^2y^2-h^2y^2+aby^2=0$$ $$(ax)^2+2.ax.hy+(hy)^2-(h^2-ab)y^2=0$$ $$(ax+hy)^2-(\sqrt {h^2-ab} y)^2=0$$ Then, $$ax=\left(h \pm \sqrt {h^2-ab}\right)y$$

I got stuck at here. Please help me to continue.

Thomas Andrews
  • 177,126
pi-π
  • 7,416

1 Answers1

1

This is the straightforward (and hence the longest) method of finding the required solution.

Continuing from your work, we get the two lines $L_1: y = m_1x$ and $L_2: y = m_2x$; where $m_1$ and $m_2$ are functions of a, h, and b.

One corner of the //gm is O(0, 0).

Since the given diagonal cuts $L_1$ and $L_2$ at A and B respectively, the co-ordinates of them can then be found. From those, obtain the co-ordinates of M, the midpoint of AB.

OA is the other diagonal.

There must be some other simpler ways. One probably involves the manipulation of sum and product of roots.

Mick
  • 17,141