Question

If the lines represented by the equation $2x^2-3xy+y^2=0$ make angles $\alpha$ and $\beta$ with $x-$axis, then $co{t}^2\alpha +co{t}^2\beta$

• A. $0$
• B. $\frac{3}{2}$
• C. $\frac{7}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is D

$\frac{5}{4}$

Explanation for the correct option:

Step 1: Drive the relation between the slopes and the coefficients of the given question of the pair of lines.

In the question, a pair of straight lines $2x^2-3xy+y^2=0$ which make angles $\alpha$ and $\beta$ with $x-$axis is given.

We know that, if a line makes the angle $\theta$ with $x-$axis, then the slope $m$ of that line can be given by: $m=\tan \theta$.

We know that if the slopes of lines in a pair of straight lines $a{x}^2+2hxy+b{y}^2=0$ are $m_1$ and $m_2$, then the relations between the slopes and the coefficients are: $m_1+m_2=-\frac{2h}{b}$ and $m_1·m_2=\frac{a}{b}$.

Therefore, the slope $m_1$ of one line is $\tan \alpha$ and the slope $m_2$ of the other line is $m_2=\tan \beta$.

And the relations between the slopes and the coefficients of the given question of the pair of lines are:

$$\begin{gathered} \tan \alpha +\tan \beta =-\frac{(-3)}{1} \\ \Rightarrow \tan \alpha +\tan \beta =3...\left(1\right) \end{gathered}$$

And

$$\begin{gathered} \tan \alpha ·\tan \beta =\frac{2}{1} \\ \Rightarrow \tan \alpha ·\tan \beta =2...\left(2\right) \end{gathered}$$

Step 2: Find the value of the given expression.

Simplify the given expression as follows:

We know that, $cot\theta =\frac{1}{\tan \theta }$.

$$\begin{gathered} co{t}^2\alpha +co{t}^2\beta =\frac{1}{{\tan }^2\alpha }+\frac{1}{{\tan }^2\beta } \\ \Rightarrow co{t}^2\alpha +co{t}^2\beta =\frac{{\tan }^2\alpha +{\tan }^2\beta }{{\tan }^2\alpha ·{\tan }^2\beta } \\ \Rightarrow co{t}^2\alpha +co{t}^2\beta =\frac{{\tan }^2\alpha +{\tan }^2\beta +2\tan \alpha ·\tan \beta -2\tan \alpha ·\tan \beta }{{\tan }^2\alpha ·{\tan }^2\beta }\{\text{Add and subtract}\left(2\tan \alpha ·\tan \beta \right)\text{from the numerator}\} \\ \Rightarrow co{t}^2\alpha +co{t}^2\beta =\frac{{\left(\tan \alpha +\tan \beta \right)}^2-2\tan \alpha ·\tan \beta }{{\tan }^2\alpha ·{\tan }^2\beta }...\left(3\right) \end{gathered}$$

Substitute the values from the equation $\left(1\right)$ and equation $\left(2\right)$ in the equation $\left(3\right)$.

$$\begin{gathered} co{t}^2\alpha +co{t}^2\beta =\frac{{\left(3\right)}^2-2\left(2\right)}{2^2} \\ \Rightarrow co{t}^2\alpha +co{t}^2\beta =\frac{9-4}{4} \\ \Rightarrow co{t}^2\alpha +co{t}^2\beta =\frac{5}{4} \end{gathered}$$

Therefore, the value of the given expression is $\frac{5}{4}$.

Hence, option (D) is the correct answer.