4.5 Q-7

Question Statement

Find the joint equation of the lines passing through the origin and perpendicular to the given equation:

$$x^2 - 2xy \tan a - y^2 = 0$$

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Background and Explanation

This problem involves finding the equations of lines that satisfy two key conditions:

1. The lines pass through the origin.
2. They are perpendicular to the given equation, which represents two intersecting lines.

To solve this, we’ll use the concept of slopes and the property that the product of the slopes of perpendicular lines is $-1$. We’ll reformulate the given equation into a more workable form and derive the required conditions for perpendicularity.

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Solution

Step 1: Simplify the Given Equation

The equation is written as:

$$x^2 - 2xy \tan a - y^2 = 0$$

We rewrite it in terms of $\frac{y}{x}$:

$$\frac{y^2}{x^2} - 2 \frac{y}{x} \tan a + 1 = 0$$

Let $\frac{y}{x} = m$. This simplifies the equation into a quadratic form:

$$m^2 - 2m \tan a + 1 = 0$$

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Step 2: Solve for $m$

Using the quadratic formula:

$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, $a = 1$, $b = -2\tan a$, and $c = 1$. Substituting these values:

$$m = \frac{2\tan a \pm \sqrt{(2\tan a)^2 - 4(1)(1)}}{2}$$

Simplify further:

$$m = \tan a \pm \sqrt{\sec^2 a}$$

Since $\sec^2 a = 1 + \tan^2 a$, the roots are:

$$m = \tan a \pm 1$$

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Step 3: Find Slopes of Perpendicular Lines

The slopes of the original lines are $m_1 = \tan a + 1$ and $m_2 = \tan a - 1$. For perpendicular lines, their slopes are the negative reciprocals:

• Slope $m_3 = \frac{\cos a}{-\sin a + 1}$
• Slope $m_4 = \frac{\cos a}{-\sin a - 1}$

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Step 4: Equations of Perpendicular Lines

The equations of lines passing through the origin are:

1. $y = m_3 x$ Substituting $m_3$:

$$y = \frac{\cos a}{-\sin a + 1} x$$

Rearrange:

$$(-\sin a + 1)y = \cos a , x$$

2. $y = m_4 x$ Substituting $m_4$:

$$y = \frac{\cos a}{-\sin a - 1} x$$

Rearrange:

$$(-\sin a - 1)y = \cos a , x$$

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Step 5: Joint Equation of the Two Lines

Combine the two equations:

$$[(-\sin a + 1)y - \cos a , x] \cdot [(-\sin a - 1)y - \cos a , x] = 0$$

Expand and simplify:

$$\cos^2 a , x^2 - 2\cos a \sin a , x y - (\cos^2 a)y^2 = 0$$

Divide through by $\cos^2 a$:

$$x^2 - 2 \tan a , x y - y^2 = 0$$

This is the joint equation of the required lines.

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Key Formulas or Methods Used

1. Quadratic formula:

$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

2. Slope of perpendicular lines: If $m$ is the slope of a line, the slope of a line perpendicular to it is $-\frac{1}{m}$.

3. Simplification of trigonometric identities: $\sec^2 a = 1 + \tan^2 a$.

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Summary of Steps

1. Rewrite the given equation in terms of $\frac{y}{x}$ and substitute $m = \frac{y}{x}$.
2. Solve the resulting quadratic equation for $m$.
3. Derive the slopes of perpendicular lines using the negative reciprocal rule.
4. Write the equations of these lines and combine them to form the joint equation.