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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
If the angle between the lines represented by $ax^2 + 2hxy + by^2 = 0$ is equal to the angle between the lines $2x^2 − 5xy + 3y^2 = 0$, then show that $100(h^2 − ab) = (a + b)^2$
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Answer

Let $\theta$ be the acute angle between the lines $a x^2+2 h x y+b y^2=0$.
$\tan \theta=\left|\frac{2 \sqrt{ h ^2- ab }}{ a + b }\right|$ $\ldots(i)$
Comparing the equation $2 x^2-5 x y+3 y^2=0$ with $a x^2+2 h x y+b y^2=0$,
We get $a =2, h =-\frac{5}{2}, b =3$
Let $\alpha$ be the acute angle between the lines given by $2 x^2-5 x y+3 y^2=0$
$\therefore \tan \alpha=\left|\frac{2 \sqrt{\left(-\frac{5}{2}\right)^2-(2)(3)}}{2+3}\right|$
$\tan \alpha=\left|\frac{2 \sqrt{\frac{25}{4}-6}}{5}\right|$
$=\left|\frac{2 \sqrt{\frac{25-24}{4}}}{5}\right|$
$=\left|\frac{2 \cdot \frac{1}{2}}{5}\right|$
$\therefore \tan \alpha=\frac{1}{5}$$\ldots(ii)$
But $\theta=\alpha$ [Given]
$ \therefore \tan \theta=\tan \alpha$
$\therefore\left|\frac{2 \sqrt{ h ^2- ab }}{ a + b }\right|=\frac{1}{5} \quad \ldots . . .[\text { From (i) and (ii)] } $
By taking square of both sides, we get
$ \frac{4\left(h^2-a b\right)}{(a+b)^2}=\frac{1}{25}$
$\therefore 100\left(h^2-a b\right)=(a+b)^2 $

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