Question

I : lf the lines $p{x}^2-qxy-y^2=0$ make angles $\alpha ,\beta$ with X-axis, then $\tan (\alpha +\beta )$ is $\left(\frac{-q}{1+p}\right)$
II : lf the lines represented by $2x^2+8xy+k{y}^2=0$ are coincident, then $k$ is $5$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

The correct option is A Only I is true

Given pair of lines
$p{x}^2-qxy-y^2=0$
Since, the angles made by the lines with the x-axis are $\alpha ,\beta$
So, let $m_1=\tan \alpha$ , $m_2=\tan \beta$
We have
$m_1+m_2=-\frac{2h}{b}$
$\Rightarrow m_1+m_2=-q$
and $m_1{m}_2=\frac{a}{b}$
$\Rightarrow m_1{m}_2=-p$
Now, $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
$=\frac{-q}{1-p}$
$\Rightarrow \tan (\alpha +\beta )=\frac{-q}{1-p}$
Hence, statement 1 is true.
Given pair of lines $2x^2+8xy+k{y}^2=0$ are coincident
$\Rightarrow h^2=ab$
$\Rightarrow 4^2=2k$
$\Rightarrow k=8$
Hence, statement 2 is false.