Question

If the angle between the lines represented by $a{x}^2+2hxy+b{y}^2=0$ is equal to the angle between line $2x^2-5xy+3y^2$, then show that $100(h^2-ab)={(a+b)}^2$.

Answer 1

Let the angle between the lines is $\theta$

Given equations are $a{x}^2+2hxy+b{y}^2=0,2x^2-5xy+3y^2=0$

we know that

$cos\theta =\frac{a+b}{\sqrt{(a-b{)}^2+4b^2}}$

For $2x^2-5xy+3y^2$

$a=2,b=3,h=\frac{-5}{2}$

so $\frac{a+b}{\sqrt{(a-b{)}^2+4b^2}}=\frac{2+3}{\sqrt{(2-3{)}^2+4(\frac{-5}{2}{)}^2}}$

$\frac{a+b}{\sqrt{(a-b{)}^2+4b^2}}=\frac{5}{\sqrt{1+25}}=\frac{5}{\sqrt{26}}$

squaring on both sides

$26(a+b{)}^2=25(a-b{)}^2+100b^2$

we know that

$(a+b{)}^2=(a-b{)}^2+4ab$

So $26(a+b{)}^2=25((a+b{)}^2-4ab)+100b^2$

$26(a+b{)}^2=25(a+b{)}^2-100ab+100b^2$

$(a+b{)}^2=100(b^2-ab)$