Question

If the two lines given by $ax^2 + 2hxy + by^2 = 0$ make inclinations $\alpha$ and $\beta$, then $\tan(\alpha + \beta) =$

• A. $\frac{h}{a+b}$
• B. $\frac{2h}{a+b}$
• C. $\frac{h}{a-b}$
• D. $\frac{2h}{a-b}$

Hint:

Don't confuse $\tan(\alpha + \beta)$ with the formula for the acute angle $\theta$ between the two lines! The angle between the lines is $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$, which relies on the difference of the slopes ($m_1 - m_2$) rather than their sum.

Answer 1

The Correct Option is D

Step 1: Understand the picture.
The equation $ax^2+2hxy+by^2=0$ is two straight lines through the origin. Their slopes are $\tan\alpha$ and $\tan\beta$, since $\alpha,\beta$ are the angles they make.
Step 2: Recall the slope rules.
For this pair of lines, the sum of slopes is $\tan\alpha+\tan\beta=-\dfrac{2h}{b}$ and the product is $\tan\alpha\tan\beta=\dfrac{a}{b}$.
Step 3: Write the formula we need.
$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.
Step 4: Put the two values in.
$\tan(\alpha+\beta)=\dfrac{-\frac{2h}{b}}{1-\frac{a}{b}}$.
Step 5: Clear the small fractions.
Multiply top and bottom by $b$: $\tan(\alpha+\beta)=\dfrac{-2h}{b-a}$.
Step 6: Tidy the sign.
Multiply top and bottom by $-1$ to write it as $\dfrac{2h}{a-b}$, which is option (4). \[ \boxed{\tan(\alpha+\beta)=\frac{2h}{a-b}} \]