Question Statement

Show that the angle between the lines represented by the following equations is a right angle:

(i) $x^2+5xy-y^2=0$

(ii) $x^2-2(\tan \theta )xy-y^2=0$

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Background and Explanation

The general second-degree homogeneous equation $a{x}^2+2hxy+b{y}^2=0$ represents a pair of straight lines passing through the origin. The angle between these lines is given by $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. When $a+b=0$, the denominator becomes zero, making $\tan \theta$ undefined (infinite), which corresponds to $\theta =90^{\circ }$.

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Solution

Part (i)

Given equation: $x^2+5xy-y^2=0$

Compare with the standard form $a{x}^2+2hxy+b{y}^2=0$:

• $a=1$
• $b=-1$
• $2h=5\Rightarrow h=\frac{5}{2}$

Let $\theta$ be the angle between the lines. Using the angle formula:

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

Substituting the identified values:

$\begin{array}{rl}\tan \theta & =\frac{2\sqrt{{\left(\frac{5}{2}\right)}^2-(1)(-1)}}{1+(-1)}\\ & =\frac{2\sqrt{\frac{25}{4}+1}}{0}\\ & =\frac{2\sqrt{\frac{29}{4}}}{0}\\ \Rightarrow \tan \theta & =\infty \end{array}$

Therefore:

$\theta ={\tan }^{-1}(\infty )=90^{\circ }$

This shows that the lines are perpendicular to each other.

Key observation: The lines are perpendicular because $a+b=1+(-1)=0$, i.e., the sum of the coefficients of $x^2$ and $y^2$ is zero.

Part (ii)

Given equation: $x^2-2(\tan \theta )xy-y^2=0$

Compare with the standard form $a{x}^2+2hxy+b{y}^2=0$:

• $a=1$
• $b=-1$
• $2h=-2\tan \theta \Rightarrow h=-\tan \theta$

Let $\alpha$ be the angle between the lines. Using the angle formula:

$\tan \alpha =\frac{2\sqrt{h^2-ab}}{a+b}$

Substituting the identified values:

$\begin{array}{rl}\tan \alpha & =\frac{2\sqrt{(-\tan \theta {)}^2-(1)(-1)}}{1+(-1)}\\ & =\frac{2\sqrt{{\tan }^2\theta +1}}{0}\\ \Rightarrow \tan \alpha & =\infty \end{array}$

Therefore:

$\alpha ={\tan }^{-1}(\infty )=90^{\circ }$

This shows that the lines are perpendicular to each other.

Key observation: The lines are perpendicular because $a+b=1+(-1)=0$, i.e., the sum of the coefficients of $x^2$ and $y^2$ is zero.

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Key Formulas or Methods Used

• Standard form of pair of lines: $a{x}^2+2hxy+b{y}^2=0$
• Angle between two lines: $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$
• Condition for perpendicular lines: $a+b=0$ (coefficient of $x^2$ + coefficient of $y^2$ = 0)
• Trigonometric identity: $1+{\tan }^2\theta ={\sec }^2\theta$ (implied in the calculation of $\sqrt{{\tan }^2\theta +1}$)

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Summary of Steps

1. Identify coefficients $a$, $b$, and $h$ by comparing the given equation with the standard form $a{x}^2+2hxy+b{y}^2=0$
2. Check the perpendicularity condition: Verify that $a+b=0$ (sum of coefficients of $x^2$ and $y^2$ equals zero)
3. Apply the angle formula $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$
4. Evaluate: Since $a+b=0$, the denominator is zero, making $\tan \theta \to \infty$
5. Conclude: $\theta ={\tan }^{-1}(\infty )=90^{\circ }$, proving the lines intersect at a right angle