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

While searching for the answers, I got this answer on 'Yahoo answers' but I did not understand anything over here. Please anyone make me clear about it.

Let the parallelogram be $OPMQ$ ($PQ$ given diagonal) then $$ax²+2hxy+by²+λ(px+qy−1)=0$$ is a conic through $P$ and $Q$. For this to be 2nd pair of sides of parallelogram it must be of form $$a(x−X)²+2h(x−X)(y−Y)+b (y−Y)²=0$$ ( that is the same as given pair of sides but translated to $M(X,Y)$ as origin ) Quadratic terms are of course identical. Comparing linear terms gives $$−2aX −2hY=λp, −2bY−2hX=λq.$$ Divide to eliminate $λ$ and get $$Y/X = (aq −hp)/(bp−hq).$$ 2nd diagonal through $O$ is $y/x=Y/X$.

Alex M.
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pi-π
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    Possible duplicate of If $ax^2+2hxy+by^2=0$ be the two sides of a $||$gm .. –  Dec 31 '16 at 07:49
  • Actually I want the explanation of the answer I gave. – pi-π Dec 31 '16 at 07:58
  • Having seen your previous answer (the reference given by @Rohan), let me say that you must recall when asking a very related question the reference to the initial question. – Jean Marie Dec 31 '16 at 08:02
  • Little explanation of the answer that has been given to you on Yahoo. It involves the concept of pencil of conics. If you have two conics $C_0$ and $C_{\infty}$ with equations $f(x,y)=0$ and $g(x,y)=0$, $f(x,y)+\lambda g(x,y)=0$ is the equation of a conic $C_{\lambda}$ (passing through the common points of $C_0$ and $C_{\infty}$). The set of all these conics is called a "pencil of conics". Pencils have many other properties. – Jean Marie Dec 31 '16 at 08:16

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Other than this answer:

Let $y = m_1x$ and $y=m_2x$ be the two lines represented by $ax^2+2hxy+by^2=0$ so that $$m_1+m_2 = -\frac{2h}{b} \text{ and } m_1m_2=\frac{a}{b}...(1)$$ Let $OA$ and $OC$ represent these lines and let the line $px+qy=1$ meets $OA$ and $OC$ at $A$ and $C$. After completing te parallelogram, we have to find the equation of the diagonal $OB$. Let the two diagonals intersect at $M(x_1, y_1)$.

The line $px+qy=1$ meets $y=m_1x$ at $A(\frac{1}{p+qm_1}, \frac{m_1}{p+qm_1})$ and $y=m_2x$ at $C(\frac{1}{p+qm_2}, \frac{m_2}{p+qm_2})$. Now the coordinates of the midpoint of $A$ and $C$ are given by $$M(x_1, y_1) = M(\frac{2p+q(m_1m_2)}{2(p+qm_1)(p+qm_2)}, \frac{p(m_1+m_2)+2qm_1m_2}{2(p+qm_1)(p+qm_2)})...(2)$$ Equation of the other diagonal $OB$ is thus, $$y-0 = \frac{y_1-0}{x_1-0}(x-0) \Rightarrow \frac{y}{x} = \frac{y_1}{x_1}$$ Substituting the coordinates from $(2)$ gives us the required answer.

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