Question

4.
The equation ax2 + 2hxy + by2 = 0 represents a pair of parallel straight lines if
(a) a = b
(b)h? = ab
(c) ab=0 (d)h = ab.
otom?answer
​

Answer 1

Given equation,

$\longrightarrow ax^2+2hxy+by^2=0$

If this equation represents a pair of straight lines, then the equations of those straight lines should be the factors of this equation.

Let us factorise it as a quadratic equation in $x,$ taking $y$ constant.

$\longrightarrow x=\dfrac{-2hy\pm\sqrt{(2hy)^2-4aby^2}}{2a}$

$\longrightarrow x=\dfrac{-2hy\pm\sqrt{4h^2y^2-4aby^2}}{2a}$

$\longrightarrow x=\dfrac{-2hy\pm2y\sqrt{h^2-ab}}{2a}$

$\longrightarrow x=\dfrac{-2y\left(h\pm\sqrt{h^2-ab}\right)}{2a}$

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

So our equation in factorised form will be,

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

And so each line, among the pair of straight lines, will be represented by,

$\longrightarrow 2ax+2y\left(h+\sqrt{h^2-ab}\right)=0\quad\quad\dots(1)$

$\longrightarrow 2ax+2y\left(h-\sqrt{h^2-ab}\right)=0\quad\quad\dots(2)$

Slope of line (1) is,

$\longrightarrow m_1=-\dfrac{2a}{2\left(h+\sqrt{h^2-ab}\right)}$

$\longrightarrow m_1=-\dfrac{a}{h+\sqrt{h^2-ab}}$

And that of line (2) is,

$\longrightarrow m_2=-\dfrac{2a}{2\left(h-\sqrt{h^2-ab}\right)}$

$\longrightarrow m_2=-\dfrac{a}{h-\sqrt{h^2-ab}}$

If these two lines are parallel, so that or equation should represent a pair of parallel straight lines, the slopes of these lines should be equal.

$\longrightarrow m_1=m_2$

$\longrightarrow -\dfrac{a}{h-\sqrt{h^2+ab}}=-\dfrac{a}{h-\sqrt{h^2-ab}}$

$\longrightarrow\underline{\underline{ab=0}}$

Hence (c) is the answer.