6.7 Q-6

Question Statement

Find the equation of the tangent to the conic $9x^2 - 4y^2 = 36$ that is parallel to the line $5x - 2y + 7 = 0$.

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Background and Explanation

To solve this problem, we need to:

1. Find the slope of the given line: The slope of the line $5x - 2y + 7 = 0$ will help us determine the slope of the tangent line that is parallel to it.
2. Find the slope of the tangent to the conic: The general equation for the slope of a tangent line to a conic can be found by differentiating the conic equation.
3. Set the slopes equal: Since the required tangent is parallel to the given line, their slopes will be the same.
4. Use the point-slope form to find the equations of the tangents at the points where the slopes match.

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Solution

Step 1: Find the slope of the given line

The given line is:

$$5x - 2y + 7 = 0$$

Rearranging the equation into slope-intercept form $y = mx + b$:

$$-2y = -5x - 7 \quad \Rightarrow \quad y = \frac{5}{2}x + \frac{7}{2}$$

Thus, the slope of the given line is $m = \frac{5}{2}$.

Step 2: Differentiate the equation of the conic to find the slope of the tangent

The equation of the conic is:

$$9x^2 - 4y^2 = 36 \tag{1}$$

Differentiate both sides with respect to $x$:

$$18x - 8y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$:

$$8y \frac{dy}{dx} = 18x \quad \Rightarrow \quad \frac{dy}{dx} = \frac{18x}{8y} = \frac{9x}{4y}$$

Thus, the slope of the tangent to the conic at any point is $\frac{9x}{4y}$.

Step 3: Set the slopes equal

Since the tangent line is parallel to the given line, their slopes must be equal. Therefore, we set the slope of the tangent equal to the slope of the line $\frac{5}{2}$:

$$\frac{9x}{4y} = \frac{5}{2}$$

Cross-multiply to solve for $x$:

$$9x = 10y \quad \Rightarrow \quad x = \frac{10y}{9} \tag{3}$$

Step 4: Substitute into the equation of the conic

Substitute $x = \frac{10y}{9}$ into the original equation $9x^2 - 4y^2 = 36$:

$$9\left(\frac{10y}{9}\right)^2 - 4y^2 = 36$$

Simplifying:

$$9 \times \frac{100y^2}{81} - 4y^2 = 36$$ $$\frac{900y^2}{81} - 4y^2 = 36$$

Multiply through by 81 to eliminate the denominator:

$$900y^2 - 324y^2 = 36 \times 81$$ $$576y^2 = 2916$$ $$y^2 = \frac{2916}{576} = \frac{81}{16}$$

Thus, $y = \pm \frac{9}{4}$.

Step 5: Find the corresponding $x$-coordinates

Substitute $y = \frac{9}{4}$ into $x = \frac{10y}{9}$:

$$x = \frac{10 \times \frac{9}{4}}{9} = \frac{10}{4} = \frac{5}{2}$$

Similarly, for $y = -\frac{9}{4}$:

$$x = \frac{10 \times \left(-\frac{9}{4}\right)}{9} = -\frac{5}{2}$$

Thus, the two points where the tangents are parallel to the given line are:

$$\left( \frac{5}{2}, \frac{9}{4} \right) \quad \text{and} \quad \left( -\frac{5}{2}, -\frac{9}{4} \right)$$

Step 6: Find the equations of the tangents

For the point $\left( \frac{5}{2}, \frac{9}{4} \right)$, the slope of the tangent is $\frac{5}{2}$. Using the point-slope form of the equation of a line:

$$y - \frac{9}{4} = \frac{5}{2} \left( x - \frac{5}{2} \right)$$

Simplifying:

$$y - \frac{9}{4} = \frac{5}{2}x - \frac{25}{4}$$ $$y = \frac{5}{2}x - \frac{25}{4} + \frac{9}{4}$$ $$y = \frac{5}{2}x - \frac{16}{4} = \frac{5}{2}x - 4$$

Multiplying through by 2 to eliminate the fraction:

$$2y = 5x - 8 \quad \Rightarrow \quad 5x - 2y - 8 = 0$$

For the point $\left( -\frac{5}{2}, -\frac{9}{4} \right)$, using the same process:

$$y + \frac{9}{4} = \frac{5}{2} \left( x + \frac{5}{2} \right)$$

Simplifying:

$$y + \frac{9}{4} = \frac{5}{2}x + \frac{25}{4}$$ $$y = \frac{5}{2}x + \frac{25}{4} - \frac{9}{4}$$ $$y = \frac{5}{2}x + \frac{16}{4} = \frac{5}{2}x + 4$$

Multiplying through by 2:

$$2y = 5x + 8 \quad \Rightarrow \quad 5x - 2y + 8 = 0$$

Thus, the two equations of the tangents are:

$$5x - 2y - 8 = 0 \quad \text{and} \quad 5x - 2y + 8 = 0$$

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Key Formulas or Methods Used

1. Slope of the Tangent to a Conic: For a conic of the form $Ax^2 + By^2 = C$, the slope of the tangent is:

$$\frac{dy}{dx} = \frac{9x}{4y}$$

2. Point-Slope Form of a Line: The equation of a line through a point $(x_1, y_1)$ with slope $m$ is:

$$y - y_1 = m(x - x_1)$$

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Summary of Steps

1. Find the slope of the given line by rewriting it in slope-intercept form.
2. Differentiate the equation of the conic to find the slope of the tangent.
3. Set the slope of the tangent equal to the slope of the given line.
4. Solve for $x$ and $y$ using the conic equation.
5. Use the point-slope form of the line to find the equations of the tangents at the points.
6. Simplify the equations into standard form.