Q.Find the combined equation of the lines joining the origin to the points where the line $x - 2y + 4 = 0$ meets the parabola $y^2 = 4x$, and hence show that the line touches the parabola.

Homogenizing and setting (coeff of $x^2$) + (coeff of $y^2$) $=0$ gives $k^2=1$, so $k=\pm1$.

Write the line as $\dfrac{x+2y}{k}=1$. Homogenize the curve $2x^2-2xy+3y^2+2x-y-1=0$ by replacing the linear and constant terms using this factor of $1$:

$2x^2 - 2xy + 3y^2 + (2x - y)\frac{x+2y}{k} - \left(\frac{x+2y}{k}\right)^2 = 0.$

Multiply by $k^2$:

$2k^2x^2 - 2k^2xy + 3k^2y^2 + k(2x - y)(x+2y) - (x+2y)^2 = 0.$

Now $(2x-y)(x+2y) = 2x^2 + 3xy - 2y^2$ and $(x+2y)^2 = x^2 + 4xy + 4y^2$. Collect the $x^2$ and $y^2$ coefficients: …

Making the curve's equation homogeneous of degree 2 using the line equation gives the pair of lines joining the origin to the intersection points; the condition "coefficient of $x^2$ + coefficient of $y^2$ = 0" then proves perpendicularity.

Curve: $x^2-xy+y^2+3x+3y-2=0$. Line: $x-y-\sqrt2=0$, i.e. $\dfrac{x-y}{\sqrt2}=1$.

To get the homogeneous (degree 2) equation of the lines joining the origin to the points where the curve meets the line, multiply the degree-1 terms by $\dfrac{x-y}{\sqrt2}$ (which equals 1 on the line) and the constant term by $\left(\dfrac{x-y}{\sqrt2}\right)^2$:

$x^2-xy+y^2 + (3x+3y)\left(\dfrac{x-y}{\sqrt2}\right) - 2\left(\dfrac{x-y}{\sqrt2}\right)^2 = 0$

Multiply throughout by $\sqrt2$ to clear denominators:

$\sqrt2(x^2-xy+y^2) + (3x+3y)(x-y) - \sqrt2(x-y)^2 = 0$

Expand $(3x+3y)(x-y) = 3x^2-3xy+3xy-3y^2 = 3x^2-3y^2$

Expand $\sqrt2(x-y)^2 = \sqrt2(x^2-2xy+y^2) = \sqrt2x^2-2\sqrt2xy+\sqrt2y^2$

Substitute and collect terms:

$\sqrt2x^2-\sqrt2xy+\sqrt2y^2 + 3x^2-3y^2 - \sqrt2x^2+2\sqrt2xy-\sqrt2y^2 = 0$

…

Homogenise the circle's equation using the chord (make every term degree 2 in $x,y$ via $lx+my=1$) to get the pair of lines joining the origin to the chord's endpoints; then impose the standard perpendicularity condition (coefficient of $x^2$ + coefficient of $y^2$ = 0).

Circle: $x^2+y^2=a^2$. Chord: $lx+my=1$.

Homogenise the circle equation using the chord (which equals $1$):

$x^2+y^2 = a^2(lx+my)^2$

$x^2+y^2 - a^2(l^2x^2+2lmxy+m^2y^2) = 0$

$(1-a^2l^2)x^2 - 2a^2lm\,xy + (1-a^2m^2)y^2 = 0$

…

Homogenising gives coefficients $x^2:(2k^2+2k-1)$ and $y^2:(3k^2-2k-4)$; perpendicularity needs their sum $=0$, i.e. $5k^2-5=0$.

The lines joining the origin to the intersection points come from homogenising the curve using $\dfrac{x+2y}{k}=1$:

$2x^2-2xy+3y^2+(2x-y)\dfrac{x+2y}{k}-\left(\dfrac{x+2y}{k}\right)^2=0$.

Multiply by $k^2$:

$2k^2x^2-2k^2xy+3k^2y^2+k(2x-y)(x+2y)-(x+2y)^2=0$.

Now $(2x-y)(x+2y)=2x^2+3xy-2y^2$ and $(x+2y)^2=x^2+4xy+4y^2$. Collect:

• coefficient of $x^2$: $2k^2+2k-1$, …

Homogenize the curve's equation using the line's equation to get the pair of lines through the origin joining it to the intersection points, then check the perpendicularity condition (coeff. of $x^2$ + coeff. of $y^2$ = 0).

Curve: $x^2-xy+y^2+3x+3y-2=0$. Line: $x-y-\sqrt2=0 \implies \dfrac{x-y}{\sqrt2}=1$.

Make the curve's equation homogeneous of degree 2 by using $1 = \dfrac{x-y}{\sqrt2}$ to raise the degree of each lower-degree term:

$x^2-xy+y^2 + (3x+3y)\left(\dfrac{x-y}{\sqrt2}\right) - 2\left(\dfrac{x-y}{\sqrt2}\right)^2 = 0$

Expand each part:

$(3x+3y)\left(\dfrac{x-y}{\sqrt2}\right) = \dfrac{3}{\sqrt2}(x+y)(x-y) = \dfrac{3}{\sqrt2}(x^2-y^2)$

$-2\left(\dfrac{x-y}{\sqrt2}\right)^2 = -2\cdot\dfrac{(x-y)^2}{2} = -(x-y)^2 = -(x^2-2xy+y^2)$

So the homogeneous equation becomes:

$x^2-xy+y^2 + \dfrac{3}{\sqrt2}x^2 - \dfrac{3}{\sqrt2}y^2 - x^2+2xy-y^2 = 0$

Collecting terms:

$x^2$ terms: $x^2+\dfrac{3}{\sqrt2}x^2-x^2 = \dfrac{3}{\sqrt2}x^2$

$y^2$ terms: $y^2-\dfrac{3}{\sqrt2}y^2-y^2 = -\dfrac{3}{\sqrt2}y^2$

$xy$ terms: $-xy+2xy = xy$

So: $\dfrac{3}{\sqrt2}x^2 + xy - \dfrac{3}{\sqrt2}y^2 = 0$

…

Homogenize the curve using the line (to get the pair of lines OA, OB through the origin), then apply the perpendicularity condition 'coefficient of $x^2$ + coefficient of $y^2$ = 0'.

Given curve: $2x^{2}-2xy+3y^{2}+2x-y+1=0$ and line: $x+2y=k$, i.e. $\dfrac{x+2y}{k}=1$.

Homogenizing: replace the constant '1' in the curve equation using $\dfrac{x+2y}{k}$ for the linear terms and $\left(\dfrac{x+2y}{k}\right)^{2}$ for the constant term, so every term becomes degree 2 (this gives the combined equation of the two lines $OA$, $OB$ joining the origin to the points where the curve meets the given line):

$2x^{2}-2xy+3y^{2} + (2x-y)\left(\frac{x+2y}{k}\right) + \left(\frac{x+2y}{k}\right)^{2} = 0$

Multiply through by $k^{2}$:

$k^{2}(2x^{2}-2xy+3y^{2}) + k(2x-y)(x+2y) + (x+2y)^{2} = 0$

Expand $(2x-y)(x+2y)=2x^{2}+3xy-2y^{2}$ and $(x+2y)^{2}=x^{2}+4xy+4y^{2}$:

$k^{2}(2x^{2}-2xy+3y^{2}) + k(2x^{2}+3xy-2y^{2}) + (x^{2}+4xy+4y^{2}) = 0$

Collecting coefficients of $x^2$ and $y^2$:

$\text{coeff}(x^{2}) = 2k^{2}+2k+1, \qquad \text{coeff}(y^{2}) = 3k^{2}-2k+4$

Perpendicularity condition for a pair of lines $Ax^{2}+2Hxy+By^{2}=0$ through the origin is $A+B=0$:

$(2k^{2}+2k+1)+(3k^{2}-2k+4) = 0$

$5k^{2}+5 = 0$

$k^{2} = -1$

…

Homogenizing the curve using the line reduces it to the joint equation of the two lines through the origin; the coefficient sum ($x^2$ coeff $+\,y^2$ coeff $=0$) proves perpendicularity.

Concept: Homogenization

To find the pair of lines joining the origin to the points where a line meets a curve, make the curve's equation homogeneous of degree 2 using the line (written as $=1$). Two lines $Ax^2+2Hxy+By^2=0$ through the origin are perpendicular iff $A+B=0$.

Step 1: Write the line as (expr) = 1

$x-y-\sqrt2=0 \Rightarrow \dfrac{x-y}{\sqrt2}=1$

Step 2: Homogenize the curve

$x^2-xy+y^2+3x+3y-2=0$ — multiply the degree-1 terms by $\dfrac{x-y}{\sqrt2}$ and the constant by $\left(\dfrac{x-y}{\sqrt2}\right)^2$:

$x^2-xy+y^2 + (3x+3y)\!\left(\dfrac{x-y}{\sqrt2}\right) - 2\!\left(\dfrac{x-y}{\sqrt2}\right)^2 = 0$

Step 3: Simplify each piece

$(3x+3y)\dfrac{x-y}{\sqrt2} = \dfrac{3(x^2-y^2)}{\sqrt2}$

$2\left(\dfrac{x-y}{\sqrt2}\right)^2 = 2\cdot\dfrac{(x-y)^2}{2} = (x-y)^2 = x^2-2xy+y^2$

So the equation becomes:

$x^2-xy+y^2+\dfrac{3(x^2-y^2)}{\sqrt2} - (x^2-2xy+y^2) = 0$

Step 4: Combine like terms

…