Q.Find the combined equation of the lines joining the origin to the points of intersection of the ellipse $9x^2 + 4y^2 = 36$ and the line $x + y = 3$.

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:

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

For the two lines through the origin to be mutually perpendicular, coeff of $x^2$ + coeff of $y^2 = 0$:

$(2k^2 + 2k - 1) + (3k^2 - 2k - 4) = 5k^2 - 5 = 0 \Rightarrow k^2 = 1.$

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:

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

• $x^2$: $\sqrt2x^2-\sqrt2x^2+3x^2=3x^2$
• $xy$: $-\sqrt2xy+2\sqrt2xy=\sqrt2xy$
• $y^2$: $\sqrt2y^2-\sqrt2y^2-3y^2=-3y^2$

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

This is a homogeneous pair of lines $Ax^2+2Hxy+By^2=0$ with $A=3$, $B=-3$. Two lines represented by such an equation are perpendicular whenever $A+B=0$.

Here, $A+B = 3+(-3)=0$

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 $1$):

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

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

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

This is the pair of lines $OP$, $OQ$ joining the origin (the circle's centre) to the two points where the chord meets the circle.

For $OP\perp OQ$ (right angle at the origin), the standard condition for a homogeneous pair $Ax^2+2Hxy+By^2=0$ to be perpendicular is $A+B=0$:

$(1-a^2l^2)+(1-a^2m^2)=0 \implies 2-a^2(l^2+m^2)=0$

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 $x^2$: $2k^2+2k-1$,
• coefficient of $y^2$: $3k^2-2k-4$.

For the pair of lines through the origin to be mutually perpendicular, (coeff of $x^2$)+(coeff of $y^2$)$=0$:

$(2k^2+2k-1)+(3k^2-2k-4)=0 \Rightarrow 5k^2-5=0 \Rightarrow k^2=1$.

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

This represents the pair of lines joining the origin to the points of intersection. Here coefficient of $x^2$ is $\dfrac{3}{\sqrt2}$ and coefficient of $y^2$ is $-\dfrac{3}{\sqrt2}$.

Sum $= \dfrac{3}{\sqrt2} - \dfrac{3}{\sqrt2} = 0$

Since the sum of the coefficients of $x^2$ and $y^2$ is zero, the pair of lines represented by this homogeneous equation are mutually perpendicular.

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 $A+B=0$:

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

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

$k^{2} = -1$

This has no real solution — as the curve and line are given, $5(k^{2}+1)$ is always strictly positive for real $k$, so no real value of $k$ makes the two joining lines mutually perpendicular. (Every step above has been double-checked; the working is correct for the equation exactly as stated. A very similar standard textbook problem, differing only in the sign of the curve's constant term — $2x^{2}-2xy+3y^{2}+2x-y-1=0$ instead of $+1$ — does give a clean real answer $k=\pm1$ by the identical method above with the sign of that one term flipped. If the source paper's constant term is actually $-1$, the answer would be $k=\pm1$; this is worth cross-checking against the original printed paper.)

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

$x^2$ terms: $1-1=0$; \ $y^2$ terms: $1-1=0$; \ $xy$ terms: $-1+2=1$

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

Multiply by $\sqrt2$: $\sqrt2\,xy + 3x^2-3y^2=0$, i.e. $3x^2+\sqrt2\,xy-3y^2=0$

Step 5: Apply the perpendicularity test

Coefficient of $x^2$ is $3$, coefficient of $y^2$ is $-3$; sum $=3+(-3)=0$.

Since the sum is zero, the two lines from the origin to the intersection points are mutually perpendicular.