Lines Joining the Origin to the Intersection of a Line and a Curve

Let us find the equation to the pair of lines joining the origin to the points of intersection of the line \[lx+my=n\] and the curve \[ax^2+2hxy+by^2+2gx+2fy+c=0\] The general equation of second degree \[ax^2+2hxy+by^2+2gx+2fy+c=0\text{ __(1)}\] in general, represents a curve, except in the case when it represents a pair of straight lines.

Lines Joining the Origin to the Intersection of a Line and a Curve

Let the straight line $lx+my=n$ meet the curve (or pair of straight lines) in two points $A$ and $B$. We have to find the equation representing the line pair $OA$ and $OB$.

The equation of given straight line may be written as \[\frac{lx+my}{n}=1\text{ __(2)}\]

Now, consider the homogeneous equation of second degree in $x$ and $y$, \[ax^2+2hxy+by^2+2(gx+fy)\left(\frac{lx+my}{n}\right)\]\[+c\left(\frac{lx+my}{n}\right)^2=0\text{ __(3)}\]

This equation, being homogeneous in $x$ and $y$, represents a pair of straight lines through the origin.

Also the coordinates of the point of intersection of the straight line $\text{(2)}$ and the curve $\text{(1)}$ satisfy both the equations $\text{(1)}$ and $\text{(2)}$, and hence they satisfy the equation $\text{(3)}$. The equation $\text{(3)}$ therefore represents a pair of lines which passes through the origin and the points of intersection of the line $\text{(2)}$ and the curve S\text{(1)}$.


Find the equation of the lines joining the origin to the points of intersection of $x+2y=3$ and $4x^2+16xy-12y^2-8x+12y-3=0$. Also find the angle between the two lines.

Here, equation of curve is, \[4x^2+16xy-12y^2-8x+12y-3=0\] and, equation of straight line is, \[x+2y=3\] \[\text{or }\frac{x+2y}{3}=1\]

Making the curve homogeneous with the help of straight line, \[4x^2+16xy-12y^2-8x.1+12y.1+3.1=0\] \[\text{or, }4x^2+16xy-12y^2-8x\left(\frac{x+2y}{3}\right)\]\[+12y\left(\frac{x+2y}{3}\right)-3\left(\frac{x+2y}{3}\right)^2=0\] \[\text{or, }12x^2+48xy-36y^2-8x^2-16xy+12xy\]\[+24y^2-x^2-4xy-4y^2=0\] \[\therefore 3x^2+40xy-16y^2=0\]

This is the equation of the lines joining the origin and the points of intersection of the given curve and the line.

Now, \[\begin{array}{c}a=3,&2h=40&\text{and}&b=-16\\a=3,&h=20&\text{and}&b=-16\end{array}\]

Let $\theta$ be the angle between the line pair. Then, \[\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[=\pm\frac{2\sqrt{400+3×16}}{3-16}=\pm\frac{16\sqrt{7}}{13}\]


Find the equation to the pair of straight lines joining the origin to the intersection of the straight line $y=mx+c$ and the curve $x^2+y^2=a^2$. Prove that they are at right angles if $2c^2=a^2(1+m^2)$.

Here, equation of curve is, \[x^2+y^2=a^2\] and, equation of straight line is, \[y=mx+c\] \[\text{i.e. }\frac{y-mx}{c}=1\]

Making the curve homogeneous with the help of straight line, we have, \[x^2+y^2=a^2.1\] \[x^2+y^2=a^2\left(\frac{y-mx}{c}\right)^2\] \[c^2x^2+c^2y^2=a^2(y^2-2mxy+m^2x^2)\] \[c^2x^2+c^2y^2=a^2(y^2-2mxy+m^2x^2)\] \[c^2x^2+c^2y^2=a^2y^2-2a^2mxy+a^2m^2x^2\] \[(c^2-a^2m^2)x^2+2a^2mxy+(c^2-a^2)y^2=0\]

This is the equation of the lines joining the origin and the points of intersection of the given curve and the line.

Now, the two lines will be right angles if \[\text{coeff. of }x^2+\text{coeff. of }y^2=0\] \[c^2-a^2m^2+c^2-a^2=0\] \[\therefore 2c^2=a^2(1+m^2)\]



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