Question

If the angle between lines is represented by an equation ${\mathrm{y}}^2+\mathrm{kxy}-{\mathrm{x}}^2{\tan }^2\mathrm{A}=0\mathrm{is}2\mathrm{A}$ then$k=$

• A. $0$
• B. $1$
• C. $2$
• D. $\tan A$

Answer 1

The correct option is A

$0$

Explanation for Correct Option:

Step 1: Apply the General formula

General equation $a{x}^2+2hxy+b{y}^2=0$

The angle between the two lines $\Rightarrow \tan \alpha =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Given equation $y^2+kxy–x^2{\tan }^{2}A=0$ ..(i)

Given angle is 2A

Step 2:Comparing (i) with the general equation, we get,

$\mathrm{a}=-{\tan }^2\mathrm{A},\mathrm{b}=1$ and $\mathrm{h}=\frac{\mathrm{k}}{2}$

Now,

$$\begin{gathered} \Rightarrow \tan 2A=\left|\frac{2\sqrt{{\displaystyle \frac{k^2}{4}+{\tan }^{2}A}}}{1-{\tan }^{2}A}\right| \\ \Rightarrow \frac{2\tan A}{\left(1-{\tan }^{2}A\right)}=\left|\frac{2\sqrt{{\displaystyle \frac{k^2}{4}+{\tan }^{2}A}}}{1-{\tan }^{2}A}\right|\left[\because \tan 2A=\frac{2\tan A}{\left(1-{\tan }^{2}A\right)}\right] \\ \Rightarrow {\tan }^{2}A=\frac{k^2}{4}+{\tan }^{2}A \\ \Rightarrow k=0 \end{gathered}$$

Hence, option (A) is the Correct