Question

The angle between the lines represented by the equation $a{x}^2+xy+b{y}^2=0$ will be $45°$, if

• A. $a=1,b=6$
• B. $a=1,b=-6$
• C. $a=6,b=1$
• D. None of these

Answer 1

The correct option is B

$a=1,b=-6$

Explanation for correct option:

Finding the values of $a,b$:

Given the equation $a{x}^2+xy+b{y}^2=0$

Now comparing the general form of an equation of a pair of line $a{x}^2+2hxy+b{y}^2=0$

We get $a=a$,$b=b$,$2h=1\Rightarrow h=\frac{1}{2}$

And the given angle is $45°$

Now the angle between the lines is given by

$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Substituting $h=\frac{1}{2}$ and $\theta =45°$in the above equation we get,

$\begin{array}{rcl}\tan 45°& =& \left|\frac{2\sqrt{{\left({\displaystyle \frac{1}{2}}\right)}^2-ab}}{a+b}\right|\\ 1& =& \left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-ab}}{a+b}\right|\to \left(i\right)\left[\because \tan 45°=1\right]\end{array}$

Let us check with the options

Option (A): $a=1,b=6$

Substitute $a=1,b=6$ in equation $\left(i\right)$

$$\begin{gathered} L.H.S=1 \\ R.H.S=\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-ab}}{a+b}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-\left(1\times 6\right)}}{1+6}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1-24}{4}}\right)}}{7}\right| \\ =\left|\frac{2}{7}\times \frac{\sqrt{-23}}{2}\right| \\ =\left|\frac{\sqrt{-23}}{7}\right| \\ L.H.S\ne R.H.S \end{gathered}$$

Therefore, option(A) is not correct

Option (B): $a=1,b=-6$

Substitute $a=1,b=6$ in equation $\left(i\right)$

$$\begin{gathered} L.H.S=1 \\ R.H.S=\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-ab}}{a+b}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-1\times -6}}{1-6}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1+24}{4}}\right)}}{-5}\right| \\ =\left|\frac{2}{-5}\times \frac{\sqrt{25}}{2}\right| \\ =\left|\frac{5}{-5}\right| \\ =\left|-1\right| \\ =1 \\ L.H.S=R.H.S \end{gathered}$$

Therefore, option(B) is not correct

Option (C): $a=6,b=1$

Substitute $a=6,b=1$ in equation $\left(i\right)$

$$\begin{gathered} L.H.S=1 \\ R.H.S=\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-ab}}{a+b}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1}{4}}\right)-\left(1\times 6\right)}}{1-6}\right| \\ =\left|\frac{2\sqrt{\left({\displaystyle \frac{1-24}{4}}\right)}}{-5}\right| \\ =\left|\frac{2}{-5}\times \frac{\sqrt{-23}}{2}\right| \\ =\left|\frac{\sqrt{-23}}{-5}\right| \\ L.H.S\ne R.H.S \end{gathered}$$

Therefore, option(C) is not correct

Option (D): None of these

This is not correct because option (B) satisfies the condition.

Therefore, only option (B) satisfies the given condition.

Hence, option (B) is the correct answer.