6.7 Q-5

Question Statement

Find the equation of the tangent to the ellipse $\frac{x^2}{4} + y^2 = 1$ which is parallel to the line $2x - 4y + 5 = 0$.

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Background and Explanation

To solve this problem, we need to find the equations of the tangents to the ellipse that are parallel to the given line. The steps involved are:

1. Find the slope of the given line: The slope of the line $2x - 4y + 5 = 0$ will help us identify the slope of the tangents that are parallel to it.

2. Use the tangent equation formula for the ellipse: The general equation for tangents to an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

$$y = mx \pm \sqrt{a^2 m^2 + b^2}$$

where $m$ is the slope of the tangent.

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Solution

Step 1: Find the slope of the given line

The given line is:

$$2x - 4y + 5 = 0$$

To find the slope, we first rewrite it in slope-intercept form $y = mx + b$:

$$-4y = -2x - 5 \quad \Rightarrow \quad y = \frac{1}{2}x + \frac{5}{4}$$

So, the slope of the line is $m = \frac{1}{2}$.

Step 2: Use the tangent equation formula

The formula for the equation of the tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

$$y = mx \pm \sqrt{a^2 m^2 + b^2}$$

For the ellipse $\frac{x^2}{4} + y^2 = 1$, we have $a^2 = 4$ and $b^2 = 1$. So the equation of the tangent becomes:

$$y = mx \pm \sqrt{4m^2 + 1}$$

Step 3: Substitute the slope $m = \frac{1}{2}$ into the tangent equation

Substituting $m = \frac{1}{2}$ into the tangent equation:

$$y = \frac{1}{2}x \pm \sqrt{\left(\frac{1}{2}\right)^2 4 + 1}$$

Simplifying:

$$y = \frac{1}{2}x \pm \sqrt{\frac{1}{4} \cdot 4 + 1} = \frac{1}{2}x \pm \sqrt{1 + 1} = \frac{1}{2}x \pm \sqrt{2}$$

Step 4: Write the two tangent equations

The two tangents are:

$$y = \frac{1}{2}x + \sqrt{2} \quad \text{and} \quad y = \frac{1}{2}x - \sqrt{2}$$

Multiplying both equations by 2 to eliminate the fraction:

$$2y = x + 2\sqrt{2} \quad \text{and} \quad 2y = x - 2\sqrt{2}$$

Rearranging these equations:

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

Thus, the two equations of the tangents to the ellipse are:

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

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Key Formulas or Methods Used

1. Slope of a Line: To find the slope of the given line $2x - 4y + 5 = 0$, we rewrote it in slope-intercept form $y = mx + b$.

2. Equation of the Tangent to an Ellipse: The formula for the equation of the tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

$$y = mx \pm \sqrt{a^2 m^2 + b^2}$$

3. Point-Slope Form: The equations of the tangents were simplified by multiplying by 2 to eliminate the fraction, resulting in standard linear equations.

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Summary of Steps

1. Find the slope of the given line by rewriting it in slope-intercept form.
2. Use the formula for the tangent to the ellipse $\frac{x^2}{4} + y^2 = 1$ to express the tangent equations.
3. Substitute the slope $m = \frac{1}{2}$ into the tangent equation formula.
4. Simplify the resulting equations to get the final tangent equations in standard form.