1
$\begingroup$

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

My solution is here; $ax^2+2hxy+by^2=0$ Multiplying by a and adding h^2y^2, $(ax+hy)^2=y^2(h^2-ab) ax+hy=+/- y√(h^2-ab)$

so, $ax+y√(h^2-ab) + hy=0, ax-y√(h^2-ab) + hy=0$ are the two straight lines represented by the given equation and also O(0,0) is the point of intersection of these lines. Now, How do I move further?

$\endgroup$
4
  • $\begingroup$ There is an issue as $O(0,0)$ is not on the line $px+qy=1$... $\endgroup$ – Martigan Feb 2 '16 at 17:45
  • $\begingroup$ @Martigan,ok then i understood that $px+qy=1$ does not pass through origin but the diagonal which we have to prove passes through origin. How do I relate all these to get to my answer? $\endgroup$ – Iaamuser user Feb 2 '16 at 17:54
  • $\begingroup$ Sorry I misread your text.... My bad. $\endgroup$ – Martigan Feb 2 '16 at 18:00
  • $\begingroup$ @ Martigan, what do you mean, I could not understand? $\endgroup$ – Iaamuser user Feb 2 '16 at 18:15
0
$\begingroup$

You have got the two lines $L_1$ and $L_2$:

$$x = \dfrac{-h\pm\sqrt{h^2-ab}}{a}y.$$

Next step is to find the points of intersection $(x_1,y_1)$ and $(x_2,y_2)$ of these lines with the known diagonal $D_1$ with equation $\;px+qy=1$.

Finding $(x_1,y_1):\quad$ Substitute the equation for $D_1$ into the equation for $L_1$ to give:

$$ax_1 = \left(-h+\sqrt{h^2-ab}\right)\left(1-px_1\right)/q.$$

Simplify to:

$$x_1 = \dfrac{-h+\sqrt{h^2-ab}}{aq-ph+p\sqrt{h^2-ab}},\qquad\text{and then } \qquad y_1 = \dfrac{1-px_1}{q}.$$

Finding $(x_2,y_2):\quad$ Substitute the equation for $D_1$ into the equation for $L_2$ to give:

$$ax_2 = \left(-h-\sqrt{h^2-ab}\right)\left(1-px_2\right)/q.$$

Simplify to:

$$x_2 = \dfrac{-h-\sqrt{h^2-ab}}{aq-ph-p\sqrt{h^2-ab}},\qquad\text{and then } \qquad y_2 = \dfrac{1-px_2}{q}.$$

$$\\$$

Now, the unknown diagonal passes through the origin and also the point $(x_1+x_2,\;y_1+y_2)$. So it has equation:

\begin{align} y &= \dfrac{y_1+y_2}{x_1+x_2} x \\ & \\ &= \left(\dfrac{\dfrac{1-px_1}{q} + \dfrac{1-px_2}{q}}{x_1+x_2}\right) x \\ & \\ &= \left(\dfrac{2}{q(x_1+x_2)} - \dfrac{p}{q}\right) x \\ & \\ &= \dfrac{aq-hp}{bp-hq} x \qquad\text{after substituting for $x_1,\; x_2$ and some tedious simplification.} \end{align}

$\endgroup$
2
  • $\begingroup$ @ Mick A, How did you get the value of $x$ in your first step? Please elaborate $\endgroup$ – Iaamuser user Feb 21 '16 at 2:16
  • $\begingroup$ @Iaamuseruser You got $ax\pm y\sqrt{h^2-ab}+hy=0$. I just re-arranged that to give $$x=\dfrac{-h\pm\sqrt{h^2-ab}}{a}y.$$ $\endgroup$ – Mick A Feb 21 '16 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.