Pair of Straight Lines through Origin
1.1 A homogeneous equation of second degree \[ ax^2 + 2hxy + by^2 = 0 \] represents a pair of straight lines through the origin if \[ h^2 \geq ab. \]
Proof: Solve the equation \(ax^2 + 2hxy + by^2 = 0\) as a quadratic in \(x\) to get \[ x = \left(\frac{-h \pm \sqrt{h^2-ab}}{a}\right)y, \qquad h^2 \geq ab \] which gives two lines as \[ ax + (h+\sqrt{h^2-ab})y = 0, \qquad ax + (h-\sqrt{h^2-ab})y = 0 \] Each of these lines pass through the Origin. The lines are real if \(h^2 > ab \), coincident if \(h^2 = ab \) and lines do not exist if \(h^2 < ab \).
1.2 The angle between the pair of straight lines \[ ax^2 + 2hxy + by^2 = 0 \] is given by \[ \tan\theta = \frac{2\sqrt{h^2-ab}}{|a+b|}. \]
Let individual pair of straight lines be \[ y-m_1x=0 \] and \[ y-m_2x=0. \]
Then \[ (y-m_1x)(y-m_2x)=0 \] gives \[ m_1m_2x^2-(m_1+m_2)xy+y^2=0. \]
Comparing it with \[ ax^2+2hxy+by^2=0, \] we get \[ \frac{m_1m_2}{a} = \frac{-(m_1+m_2)}{2h} = \frac{1}{b}. \]
Therefore, \[ m_1m_2=\frac{a}{b}, \qquad m_1+m_2=-\frac{2h}{b}. \]
So, \[ \tan\theta = \left| \frac{m_1-m_2}{1+m_1m_2} \right| \] and \[ \tan\theta = \frac{ \sqrt{(m_1+m_2)^2-4m_1m_2} }{ 1+m_1m_2 }. \]
Substituting \[ m_1+m_2=-\frac{2h}{b}, \qquad m_1m_2=\frac{a}{b}, \] we obtain \[ \tan\theta = \frac{ \sqrt{ \dfrac{4h^2}{b^2} - \dfrac{4a}{b} } }{ 1+\dfrac{a}{b} }. \]
Hence, \[ \boxed{ \tan\theta = \frac{2\sqrt{h^2-ab}}{a+b} } \]
✔ The condition for coincidence of lines is \(h^2 = ab\).
✔ The condition for perpendicularity of lines is \(a + b = 0\).
✔ The acute angle is given by \(\tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}\).
1.3 The equation of the bisectors of the angles betweeen pair of straight lines \[ ax^2 + 2hxy + by^2 = 0 \] is given by \[ \frac{x^2-y^2}{a-b} = \frac{xy}{h}. \]
Let the given pair of straight lines be \[ ax^2+2hxy+by^2=0. \]
Let \(\theta_1\) and \(\theta_2\) be the angles made by these lines with the \(x\)-axis. The combined equation of the lines is
\[ (y-\tan\theta_1\,x)(y-\tan\theta_2\,x)=0. \]
Expanding, \[ y^2-(\tan\theta_1+\tan\theta_2)xy +\tan\theta_1\tan\theta_2\,x^2=0. \]
Comparing this with \[ ax^2+2hxy+by^2=0, \] we have \[ \tan\theta_1\tan\theta_2=\frac{a}{b}, \qquad \tan\theta_1+\tan\theta_2=-\frac{2h}{b}. \]
Let \(\alpha,\beta\) be the angles made by the two angle bisectors with the \(x\)-axis. We have \[ \alpha=\frac{\theta_1+\theta_2}{2}, \qquad \beta=90^\circ+\frac{\theta_1+\theta_2}{2}. \]
Therefore, \[ \tan\alpha = \tan\left(\frac{\theta_1+\theta_2}{2}\right), \] and \[ \tan\beta = \tan\left(90^\circ+ \frac{\theta_1+\theta_2}{2}\right) = -\cot\left(\frac{\theta_1+\theta_2}{2}\right). \]
The combined equation of the angle bisectors is therefore \[ (y-\tan\alpha\,x)(y-\tan\beta\,x)=0. \]
Substituting the values of \(\tan\alpha\) and \(\tan\beta\), \[ \left( y-\tan\frac{\theta_1+\theta_2}{2}\,x \right) \left( y+\cot\frac{\theta_1+\theta_2}{2}\,x \right)=0. \]
Expanding, \[ y^2-x^2+ \left[ \cot\frac{\theta_1+\theta_2}{2} - \tan\frac{\theta_1+\theta_2}{2} \right]xy=0. \]
Using \[ \cot\frac{\theta_1+\theta_2}{2} - \tan\frac{\theta_1+\theta_2}{2} = \frac{ 1-\tan^2\frac{\theta_1+\theta_2}{2} }{ \tan\frac{\theta_1+\theta_2}{2} }, \] we get \[ y^2-x^2+ \left[ \frac{ 1-\tan^2\frac{\theta_1+\theta_2}{2} }{ \tan\frac{\theta_1+\theta_2}{2} } \right]xy=0. \]
Using the double-angle formula, \[ \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} \implies \frac{1-\tan^2\theta}{\tan\theta} = \frac{2}{\tan 2\theta} \]
we have \[ y^2-x^2+ \frac{2xy}{\tan(\theta_1+\theta_2)} =0. \]
Now, \[ \tan(\theta_1+\theta_2) = \frac{ \tan\theta_1+\tan\theta_2 }{ 1-\tan\theta_1\tan\theta_2 }. \]
Substituting \[ \tan\theta_1+\tan\theta_2=-\frac{2h}{b}, \qquad \tan\theta_1\tan\theta_2=\frac{a}{b}, \] we obtain \[ \tan(\theta_1+\theta_2) = \frac{-2h/b}{1-a/b} = -\frac{2h}{a-b}. \]
Hence, \[ y^2-x^2+ \frac{h}{a-b}xy=0. \]
Therefore, \[ \boxed{ \frac{x^2-y^2}{a-b} = \frac{xy}{h} } \]
Alternate Proof:
Let individual pair of straight lines be \(y-m_1x=0\) and \(y-m_2x=0\). Since perpendicular distance from bisector on any of these two lines are equal, we have
\[ \frac{y-m_1x}{\sqrt{1+m_1^2}} = \pm \frac{y-m_2x}{\sqrt{1+m_2^2}} \]
Squaring both sides,
\[ \frac{(y-m_1x)^2}{1+m_1^2} = \frac{(y-m_2x)^2}{1+m_2^2} \]
Therefore,
\[ \begin{aligned} &(y^2+m_1^2x^2-2m_1xy)(1+m_2^2) \\ &\qquad = (y^2+m_2^2x^2-2m_2xy)(1+m_1^2). \end{aligned} \]
On simplifying,
\[ y^2(m_2^2-m_1^2) +(m_1^2-m_2^2)x^2 -2xy\left[ m_1+m_1m_2^2-m_2-m_2m_1^2 \right]=0. \]
Hence,
\[ (m_1^2-m_2^2)(x^2-y^2) -2xy(m_1-m_2)(1-m_1m_2)=0. \]
Since \[ m_1^2-m_2^2=(m_1+m_2)(m_1-m_2), \] we get
\[ (m_1+m_2)(x^2-y^2) = 2xy(1-m_1m_2). \]
Comparing \[ ax^2+2hxy+by^2=0, \] with \[ (y-m_1x)(y-m_2x)=0, \] we have \[ m_1+m_2=-\frac{2h}{b}, \qquad m_1m_2=\frac{a}{b}. \]
Substituting these values,
\[ \left(-\frac{2h}{b}\right)(x^2-y^2) = 2xy\left(\frac{b-a}{b}\right). \]
Therefore,
\[ \boxed{ \frac{x^2-y^2}{a-b} = \frac{xy}{h} } \]
Pair of Straight Lines through General Equation of Second Degree
2.1 A general equation of second degree \[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0. \] represents a pair of straight lines if and only if \[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0. \] or equivalently, \[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]
To prove the "if and only if" condition rigorously, we must treat the quadratic as a quadratic in \(x\) (or \(y\) ) and analyze its factorization.
Necessary condition:
Rewrite the equation grouping powers of \(x\): \[ ax^2+(2hy+2g)x+(by^2+2fy+c)=0 \]
For this to represent a pair of straight lines, when solved for \(x\), it must give two linear expressions in \(y\). Using the quadratic formula:
\[ x= \frac{-(2hy+2g)\pm \sqrt{(2hy+2g)^2-4a(by^2+2fy+c)}}{2a} \]
Simplify the discriminant \(D_x\) (with respect to \(x\)):
\[ D_x=(2hy+2g)^2-4a(by^2+2fy+c) \]
\[ D_x=4(h^2y^2+2hgy+g^2)-4a(by^2+2fy+c) \]
\[ D_x=4\left[ (h^2-ab)y^2+2(hg-af)y+(g^2-ac) \right] \]
For the original equation to represent two straight lines, the value of \(x\) must be linear in \(y\). This means the square root \(\sqrt{D_x} \) must be a perfect square of a linear expression in \(y\).
A quadratic in \(y\) is a perfect square if and only if its discriminant is zero. Let its discriminant be \(\Delta_y\): \[ \Delta_y = \left[2(hg-af)\right]^2 - 4(h^2-ab)(g^2-ac) =0 \] Expanding this yields \[ \boxed{ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 } \]
Sufficient condition:
Assuming the determinant is zero,
\[ abc+2fgh-af^2-bg^2-ch^2=0 \]
We know this is equivalent to \( \Delta_y=0 \). If \(\Delta_y=0\), the quadratic in \(y\) inside the square root is a perfect square.
Therefore, \[ \sqrt{D_x}=2(\alpha y+\beta), \] which is linear in \(y\).
Substitute back into the quadratic formula:
\[ x= \frac{-2hy-2g\pm2(\alpha y+\beta)}{2a} \]
\[ x= \frac{(-h+\alpha)y+(-g+\beta)}{a} \]
This gives two distinct equations of the form:
\[ x=m_1y+c_1 \qquad\text{and}\qquad x=m_2y+c_2 \]
which are clearly two straight lines.
Alternate Proof:
Let the equation \[ ax^2+2hxy+by^2+2gx+2fy+c=0 \] represent a pair of straight lines and let \((x_1,y_1)\) be their point of intersection.
If the origin is transferred to \((x_1,y_1)\), with the new axes parallel to their original directions, then \[ x\rightarrow x+x_1, \qquad y\rightarrow y+y_1. \]
Therefore, the new equation is \[ \begin{aligned} &a(x+x_1)^2+2h(x+x_1)(y+y_1) +b(y+y_1)^2\\ &\qquad+2g(x+x_1)+2f(y+y_1)+c=0. \end{aligned} \]
With respect ot new axes, this equation must be pair of straight lines through origin, and hence it must be a homogeneous quadratic equation.
On simplification, \[ \begin{aligned} &ax^2+2hxy+by^2\\ &\quad+2(ax_1+hy_1+g)x +2(hx_1+by_1+f)y\\ &\quad+ \left( ax_1^2+2hx_1y_1+by_1^2 +2gx_1+2fy_1+c \right)=0. \end{aligned} \]
Since this is a homogeneous quadratic equation, the coefficients of \(x\), \(y\), and the constant term must vanish. Thus, \[ ax_1+hy_1+g=0, \qquad ---- (1) \] \[ hx_1+by_1+f=0, \qquad ---- (2) \] \[ ax_1^2+2hx_1y_1+by_1^2 +2gx_1+2fy_1+c=0. \qquad ---- (3) \]
Multiplying (1) by \(x_1\), (2) by \(y_1\), adding them and subtracting from (3) gives \[ gx_1 + fy_1 + c = 0 \qquad ---- (4) \]
Solving (1) and (2) gives their point of intersection as \[ x_1=\frac{hf-bg}{ab-h^2}, \qquad y_1=\frac{hg-af}{ab-h^2}. \]
Substituting these values in equation (4) and simplifying gives \[ \boxed{ abc+2fgh-af^2-bg^2-ch^2=0. } \]
Yet another alternate proof:
If the equation \[ ax^2+2hxy+by^2+2gx+2fy+c=0 \] represents a pair of straight lines, then the expression on the left-hand side can be split into two linear factors: \[ lx+my+n \] and \[ l'x+m'y+n'. \]
Thus, \[ ax^2+2hxy+by^2+2gx+2fy+c = (lx+my+n)(l'x+m'y+n'). \]
Comparing coefficients on both sides, we get \[ ll'=a, \qquad mm'=b, \qquad nn'=c, \] and \[ lm'+ml'=2h, \] \[ ln'+l'n=2g, \] \[ mn'+m'n=2f. \]
Multiplying the last three equations, we get \[ (lm'+l'm)(ln'+l'n)(mn'+m'n)=8fgh. \]
Expanding, \[ \begin{aligned} &\left( l^2m'n' + ll'm'n + ll'mn' + l'^2mn \right)(mn'+m'n) =8fgh. \end{aligned} \]
On further expansion, \[ \begin{aligned} &mm'l^2n'^2 + mm'll'nn' + ll'm^2n'^2 + nn'l'^2m^2\\ &\quad nn'l^2m'^2 + ll'm'^2n^2+ ll'mm'nn' + mm'l'^2n^2 =8fgh. \end{aligned} \]
Rearranging, \[ \begin{aligned} &ll'\left(mn'^2+m'n^2\right) +mm'\left(l^2n'^2+l'^2n^2\right)\\ &\quad +nn'\left(l'^2m^2+l^2m'^2\right) +2ll'mm'nn' =8fgh. \end{aligned} \]
Now, using \[ ll'=a,\qquad mm'=b,\qquad nn'=c, \] we get \[ m^2n'^2+m'^2n^2 =(mn'+m'n)^2-2mm'nn' = 4f^2-2bc, \] \[ l^2n'^2+l'^2n^2 =(ln'+l'n)^2-2ll'nn' =4g^2-2ac, \] \[ l^2m'^2+l'^2m^2 =(lm'+l'm)^2-2ll'mm' =4h^2-2ab. \]
Therefore, our expansion becomes \[ \begin{aligned} &abc+a(4f^2-2bc) +b(4g^2-2ac) +c(4h^2-2ab) =8fgh. \end{aligned} \]
Simplifying, \[ 4af^2+4bg^2+4ch^2-4abc=8fgh. \]
Or \[ \boxed{ abc+2fgh-af^2-bg^2-ch^2=0 }. \]
2.2 To find the equation of the pair of straight lines passing through the origin and the two points of intersection of the curve \(ax^2+2hxy+by^2+2gx+2fy+c=0\) and the line \(lx + my + n = 0\)
Suppose the curve is \[ S\equiv ax^2+2hxy+by^2+2gx+2fy+c=0 \] and the line is \[ L\equiv lx+my+n=0. \]
Step 1: Make the equation of the line in the form: \(\frac{lx+my}{-n}=1 \) (i.e., Linear term = 1).
Step 2: Multiply the linear terms \(2gx\) and \(2fy\) of the curve by \(1\) (i.e., \(\frac{lx+my}{-n}\)).
Step 3: Multiply the constant term \(c\) by \(1^2\) (i.e., \(\left(\frac{lx+my}{-n}\right)^2\)).
The equation of the pair of straight lines passing through the origin and the two points of intersection of the curve and the line is \[ \boxed{ ax^2+2hxy+by^2 +(2gx+2fy)\left(\frac{lx+my}{-n}\right) +c\left(\frac{lx+my}{-n}\right)^2=0 } \]
This process is known as homogenization.
Exercises
- If the equation \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) represents a pair of straight lines, show that they intersect at the point \[ \left( \frac{hf-bg}{ab-h^2}, \frac{hg-af}{ab-h^2} \right). \]
- Show that the pair of straight lines through origin, and perpendicular to the lines \(ax^2 + 2hxy + by^2 = 0\) are given by \(bx^2 - 2hxy + ay^2 = 0\).
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Show that the four lines given by \(2x^2 + 3xy - 2y^2 = 0\) and \(2x^2 + 3xy - 2y^2 - 17x + 6y + 8 = 0\) form a rectangle. Also,
(a) Find the area of the rectangle.
(b) Find the joint equation of the diagonals.
(c) Find the point of intersection of the diagonals.
(d) Find the angle between the diagonals.
(e) Find the equation of the circle circumscribing this rectangle.