Show that every homogeneous equation of degree two in $x$ and $y$, i.e., $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines passing through origin if $h^2 - ab \ge 0$.

Show that every homogeneous equation of degree two in x and y, i.e., ax² + 2hxy + by² = 0 represents a pair of

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Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 3mImportance★★★★★

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Multiply through by $a$ and complete the square in $x$ to factor the expression into two linear factors when $h^2-ab\ge0$ .

Consider $ax^2+2hxy+by^2=0$ with $a\ne0$ .

Multiply both sides by $a$ :

$a^2x^2+2ahxy+aby^2=0$

Add and subtract $h^2y^2$ to complete the square in $x$ :

$a^2x^2+2ahxy+h^2y^2 - h^2y^2+aby^2=0$

$(ax+hy)^2=(h^2-ab)y^2$

Case $h^2-ab\ge0$ : the right side is a perfect square, $(h^2-ab)y^2=\left(\sqrt{h^2-ab}\,y\right)^2$ , so

$(ax+hy)^2-\left(\sqrt{h^2-ab}\,y\right)^2=0$

This is a difference of squares, factoring as:

$\left(ax+hy-\sqrt{h^2-ab}\,y\right)\left(ax+hy+\sqrt{h^2-ab}\,y\right)=0$

i.e.

$\left[ax+\left(h-\sqrt{h^2-ab}\right)y\right]\left[ax+\left(h+\sqrt{h^2-ab}\right)y\right]=0$

This is a product of two real linear expressions in $x,y$ with no constant term, so each factor equated to zero represents a straight line passing through the origin $(0,0)$ . Hence the original equation represents a pair of (real, possibly coincident) straight lines through the origin.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$k$
›Reveal solutionSolution

Compare with $ax^2+2hxy+by^2=0$ ; sum of slopes $=-2h/b$ , product $=a/b$ .

Compare $x^2+kxy-3y^2=0$ with $ax^2+2hxy+by^2=0$ : $a=1,\ 2h=k \Rightarrow h=\dfrac{k}{2},\ b=-3$ .

Sum of slopes $=-\dfrac{2h}{b}=-\dfrac{k}{-3}=\dfrac{k}{3}$

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CBSE 2024Set ANNUAL2 marks
$k$
›Reveal solutionSolution

Sum of slopes $=\dfrac{-2h}{b}$ , product $=\dfrac ab$ ; set sum $=2\times$ product.

For $x^2+kxy-3y^2=0$ : $a=1,\ 2h=k,\ b=-3$

Sum of slopes $=\dfrac{-2h}{b}=\dfrac{-k}{-3}=\dfrac k3$

Product of slopes $=\dfrac ab=\dfrac{1}{-3}=-\dfrac13$

…

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CBSE 2023Set ANNUAL2 marks
$ax^2+2hxy+by^2=0$
›Reveal solutionSolution

Since $h^2=ab$ , the discriminant of the slope-difference is zero, so the slopes are equal.

For $ax^2+2hxy+by^2=0$ : $m_1+m_2=\dfrac{-2h}{b}$ , $m_1m_2=\dfrac ab$ .

…

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CBSE 2022Set ANNUAL2 marks
$k$
›Reveal solutionSolution

Substitute $y=-2x$ (from $2x+y=0$ ) into the pair of lines equation; it must vanish identically in $x$ .

If $2x+y=0$ , i.e. $y=-2x$ , is one of the lines represented by $3x^2+kxy+2y^2=0$ , substituting $y=-2x$ must give $0$ for all $x$ :

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CBSE 2016Set ANNUAL2 marks
$k$
›Reveal solutionSolution
Since $2x+y=0$ is one factor, find the matching second factor of $6x^2+kxy+y^2$ and compare coefficients.

The pair $6x^2+kxy+y^2=0$ factors as $(2x+y)(cx+dy)=0$ for some constants $c,d$ (since $2x+y=0$ is given as one line).

Expanding: $(2x+y)(cx+dy)=2cx^2+(2d+c)xy+dy^2$ .

Compare with $6x^2+kxy+y^2$ :

Coefficient of $x^2$ : $2c=6 \Rightarrow c=3$

Coefficient of $y^2$ : $d=1$ …
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Q.Show that every homogeneous equation of degree two in $x$ and $y$ , i.e., $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines passing through origin if $h^2 - ab \ge 0$ .

Concept understanding — Homogeneous Pair of Lines Through the Origin

Q.Show that every homogeneous equation of degree two in xxx and yyy, i.e., ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0ax2+2hxy+by2=0 represents a pair of lines passing through origin if h2−ab≥0h^2 - ab \ge 0h2−ab≥0.

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Q.Show that every homogeneous equation of degree two in $x$ and $y$ , i.e., $ax^2 + 2hxy + by^2 = 0$ represents a pair of lines passing through origin if $h^2 - ab \ge 0$ .

Unlock everything free for 14 days