Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the co-ordinate axes if $(a+b)^2 = 4h^2$.

Prove that one of the straight lines given by ax² + 2hxy + by² = 0 will bisect the angle between the co-ordina

Exercise 6.4 · Q16

ax^2 + 2hxy + by^2 = 0

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Substitute $y=mx$ into $ax^2+2hxy+by^2=0$ to get $bm^2+2hm+a=0$ ; require $m=1$ or $m=-1$ (slopes of $y=x,y=-x$ ) to be a root, then combine both cases by squaring.

The coordinate axes are bisected by the lines $y=x$ (slope $1$ ) and $y=-x$ (slope $-1$ ). "One of the lines of the pair bisects the angle between the axes" means one of the pair's two lines is $y=x$ or $y=-x$ , i.e. has slope $m=1$ or $m=-1$ .

Step 1. Slope equation of the pair. Dividing $ax^2+2hxy+by^2=0$ by $x^2$ and putting $m=y/x$ :

bm^2+2hm+a=0.

The two roots of this quadratic in $m$ are the slopes of the two lines of the pair.

Step 2. Require $m=1$ to be a root (line is $y=x$ ). Substitute $m=1$ :

b(1)^2+2h(1)+a=0\ \Longrightarrow\ a+b+2h=0\ \Longrightarrow\ a+b=-2h.

Step 3. Require $m=-1$ to be a root (line is $y=-x$ ). Substitute $m=-1$ : …

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

CBSE 2024Set ANNUAL1 markMCQ
$x^2-4y^2=0$
(a) $\dfrac{1}{2}a^2$
(b) $2a^2$
(c) $\dfrac{2}{\sqrt3}a^2$
(d) $\dfrac{\sqrt3}{2}a^2$
›Reveal solutionSolution

The triangle formed by $x^2-4y^2=0$ and $x=a$ has area $\dfrac{1}{2}a^2$ .

Factor $x^2-4y^2=(x-2y)(x+2y)=0$ , so the two lines are $x=2y$ and $x=-2y$ , both passing through the origin $(0,0)$ .

The line $x=a$ meets $x=2y$ at $y=\dfrac{a}{2}$ , giving the point $\left(a,\dfrac{a}{2}\right)$ , and meets $x=-2y$ at $y=-\dfrac{a}{2}$ , giving $\left(a,-\dfrac{a}{2}\right)$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$6x^2 - xy + 4cy^2 = 0$
(a) 3
(b) $-3$
(c) 1
(d) $-1$
›Reveal solutionSolution

Writing $6x^2-xy+4cy^2$ as the product $(3x+4y)(ax+by)$ and matching coefficients gives $c=-3$ .

Since $3x+4y=0$ is one of the two lines represented by $6x^2-xy+4cy^2=0$ , we can write:

$6x^2-xy+4cy^2 = (3x+4y)(ax+by)$

for some constants $a,b$ (up to an overall scalar, but let's match directly). Expanding the right side:

$(3x+4y)(ax+by) = 3a\,x^2 + (3b+4a)xy + 4b\,y^2$

…

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Q.Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the co-ordinate axes if $(a+b)^2 = 4h^2$ .

Concept understanding — Pair of Straight Lines

Pair of Straight Lines

Q.Prove that one of the straight lines given by ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0ax2+2hxy+by2=0 will bisect the angle between the co-ordinate axes if (a+b)2=4h2(a+b)^2 = 4h^2(a+b)2=4h2.

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Q.Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the co-ordinate axes if $(a+b)^2 = 4h^2$ .

Unlock everything free for 14 days