Hey guys! Let's dive into a fun problem involving hyperbolas and tangents. We're going to find the equation of the tangent to the hyperbola 9x² - 4y² = 36 that's parallel to the line 8x - 15y = 1. Sounds exciting, right? This is a classic problem in conic sections, and by the end of this, you'll be able to tackle similar questions with ease. So, grab your pencils, and let's get started!
Understanding the Hyperbola
Before we jump into the solution, let's make sure we're all on the same page about hyperbolas.
Hyperbolas are fascinating curves that look like two mirrored parabolas opening away from each other. The general equation of a hyperbola centered at the origin is given by x²/a² - y²/b² = 1. This form is super important because it tells us a lot about the hyperbola’s shape and orientation. In our case, we have the equation 9x² - 4y² = 36. To get it into the standard form, we need to divide both sides by 36. Doing so, we get x²/4 - y²/9 = 1. Now we can clearly see that a² = 4 and b² = 9. This means a = 2 and b = 3. These values are crucial for understanding the hyperbola’s geometry, particularly its vertices and asymptotes.
The values 'a' and 'b' define the semi-major and semi-minor axes of the hyperbola, respectively. The vertices, which are the points where the hyperbola intersects its major axis, are located at (±a, 0). In our case, the vertices are at (±2, 0). Understanding these basics is super important because the tangent lines we're trying to find will interact with the hyperbola based on its shape and orientation. So, knowing these parameters helps us visualize and solve the problem more effectively. Moreover, this standard form allows us to easily apply formulas and techniques related to hyperbolas, such as finding the equation of tangents using derivatives or other geometric properties. By having a solid grasp of the hyperbola's characteristics, we can move forward with confidence in finding the tangent lines that meet our specific conditions.
Parallel Lines and Tangent Conditions
The key here is that the tangent line we're looking for is parallel to the line 8x - 15y = 1. Remember, parallel lines have the same slope. So, our first task is to find the slope of the given line. To do this, we can rewrite the equation in slope-intercept form (y = mx + c), where 'm' is the slope. Rearranging 8x - 15y = 1, we get 15y = 8x - 1, and further, y = (8/15)x - 1/15. Voila! The slope of this line is 8/15. This is the slope that our tangent line must also have. This concept is super important because it directly links the given condition (parallelism) to a specific mathematical property (slope). Understanding this connection allows us to translate the geometric requirement into an algebraic one, which we can then use to solve for the tangent line.
Now, let's think about the conditions for a line to be tangent to a hyperbola. A tangent line touches the hyperbola at only one point. This means that if we solve the equations of the hyperbola and the tangent line simultaneously, we should get only one solution. This condition is essential because it provides us with a way to find the point of tangency. In other words, the quadratic equation formed by substituting the equation of the line into the equation of the hyperbola should have a discriminant equal to zero. This is because a quadratic equation with a discriminant of zero has exactly one real root, which corresponds to the single point of intersection between the tangent line and the hyperbola. This understanding of tangency conditions allows us to set up an equation that we can solve for the unknown parameters of the tangent line, such as its y-intercept.
Finding the Tangent Equation
Okay, we know the slope of the tangent line is 8/15. So, the equation of our tangent line will be in the form y = (8/15)x + c, where 'c' is the y-intercept, which we need to find.
Here's where things get interesting! We're going to use the condition for tangency. We'll substitute our tangent line equation into the hyperbola equation (x²/4 - y²/9 = 1) and ensure that the resulting quadratic equation has only one solution. Substituting y = (8/15)x + c into x²/4 - y²/9 = 1, we get:
x²/4 - ((8/15)x + c)²/9 = 1
Now, let's expand and simplify this equation. It might look a bit messy, but don't worry; we'll break it down step by step. Expanding the square, we have:
x²/4 - (64x²/225 + (16/15)cx + c²)/9 = 1
Multiplying through by 9 to get rid of the fraction in the denominator of the second term:
9x²/4 - (64x²/225 + (16/15)cx + c²) = 9
Now, let's get rid of the remaining fractions by multiplying the entire equation by the least common multiple of the denominators, which is 900:
225x² - 4(64x² + 15(16/15)cx + 100c²) = 900
Simplifying further:
225x² - 256x² - 640cx - 100c² = 900
Combining like terms, we get a quadratic equation in x:
-31x² - 640cx - (100c² + 900) = 0
Or, multiplying by -1 to make the leading coefficient positive:
31x² + 640cx + (100c² + 900) = 0
Phew! That was a lot of algebra, but we're getting there. Remember, for tangency, this quadratic equation must have exactly one solution. This means its discriminant (Δ) must be equal to zero. The discriminant is given by Δ = B² - 4AC, where A = 31, B = 640c, and C = 100c² + 900. So, we have:
(640c)² - 4 * 31 * (100c² + 900) = 0
Expanding and simplifying:
409600c² - 124 * (100c² + 900) = 0
409600c² - 12400c² - 111600 = 0
397200c² = 111600
c² = 111600 / 397200
c² = 1116 / 3972
c² = 31 / 111
Taking the square root, we get:
c = ±√(31/111)
So, we have two possible values for c, which means there are two tangent lines that satisfy the condition. Now we can write the equations of the tangent lines by substituting these values of c back into y = (8/15)x + c.
The Tangent Equations
We found two values for c: c = ±√(31/111). This means we have two tangent lines parallel to 8x - 15y = 1. Let's write down their equations:
Tangent Line 1: y = (8/15)x + √(31/111)
Tangent Line 2: y = (8/15)x - √(31/111)
These are the equations of the tangents to the hyperbola 9x² - 4y² = 36 that are parallel to the line 8x - 15y = 1. We did it!
To make these equations look a bit cleaner, we can multiply through by 15 to get rid of the fraction in the slope:
Tangent Line 1: 15y = 8x + 15√(31/111)
Tangent Line 2: 15y = 8x - 15√(31/111)
And finally, we can rearrange them into the standard form:
Tangent Line 1: 8x - 15y + 15√(31/111) = 0
Tangent Line 2: 8x - 15y - 15√(31/111) = 0
These are our final answers. These two lines are tangent to the given hyperbola and parallel to the given line.
Conclusion
So, finding the equation of the tangent to a hyperbola that's parallel to a given line involves a few key steps: understanding the hyperbola's equation, using the condition for parallel lines (same slope), applying the condition for tangency (discriminant equals zero), and solving the resulting equations. It might seem like a lot, but with practice, you'll become a pro at these problems. Great job, guys! Keep up the fantastic work, and remember, math is all about breaking down complex problems into manageable steps. You've got this!
I hope this explanation was helpful and clear. If you have any more questions or want to explore other conic sections problems, feel free to ask. Happy problem-solving!