If the line αx + 2y = 1, where α ∈ ℝ, does not meet the hyperbola x² - 9y² = 9, then a possible valu

Q: Wait, why do we use the discriminant $D < 0$ here? Isn't that only for parabolas?

Great question! The discriminant is a universal tool for any quadratic equation $Ax^2 + Bx + C = 0$. Whether it comes from a parabola, a hyperbola, or a circle, the discriminant tells us about the nature of the roots. If $D < 0$, there are no real values of $x$ that satisfy the equation, meaning the line and the curve never meet in the real Cartesian plane.

Q: Do I need to worry about the case where the coefficient of $x^2$ is zero?

You have a sharp eye! If $4 - 9\alpha^2 = 0$, the equation becomes linear, representing a line parallel to the asymptote of the hyperbola. While technically a special case, in this specific problem, the inequality $D < 0$ naturally leads us to the region where the line avoids the hyperbola entirely. Always keep an eye on the leading coefficient, but here the discriminant approach covers the necessary ground.

Q: Why did we get $\alpha^2 > 5/9$? Does this mean $\alpha$ can be negative?

Absolutely! The inequality $\alpha^2 > 5/9$ implies $|\alpha| > \sqrt{5}/3$. This means $\alpha$ can be any value greater than $0.745$ or less than $-0.745$. Since our options are all positive, we focus on the positive range, but never forget that the hyperbola is symmetric!

Q: Is there a faster way using the condition of tangency $c^2 = a^2m^2 - b^2$?

Yes! You can rewrite the line as $y = -(\alpha/2)x + 1/2$. Comparing this to $y = mx + c$, you can use the tangency condition for a hyperbola. If the line doesn't intersect, it must be 'outside' the tangency limit. It's a faster method, but the discriminant method is more robust if you ever forget the formula!

Q: Why is the answer 0.8 and not 0.7?

We calculated that $|\alpha| > \sqrt{5}/3$, which is approximately $0.745$. Since $0.7$ is less than $0.745$, a line with $\alpha = 0.7$ would actually intersect the hyperbola. Only $0.8$ is greater than $0.745$, placing the line safely in the non-intersecting region.

Question

If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:

Options:

(A) 0.5

(B) 0.6

(C) 0.7

(D) 0.8

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a vast, two-dimensional plane. Before you lies a hyperbola, defined by the elegant equation: $$\frac{x^2}{9} - \frac{y^2}{1} = 1$$ It is a curve of infinite reach, stretching toward the corners of the universe, bounded by its invisible asymptotes. Now, consider a line, $\alpha x + 2y = 1$. Depending on the value of $\alpha$, this line might slice through the hyperbola like a blade, or it might dance around it, never touching its branches. Today, we are going to find the secret values of $\alpha$ that keep this line forever separated from the curve. ### The Algebraic Bridge To understand the interaction between these two geometric entities, we must bring them into the same algebraic language. We start by isolating $y$ from our line equation: $$y = \frac{1 - \alpha x}{2}$$ This is our bridge. By substituting this expression for $y$ into the hyperbola's equation, $x^2 - 9y^2 = 9$, we are essentially asking the math to tell us where they meet. If they meet, the resulting equation will have real solutions for $x$; if they do not, the math will reveal a contradiction. Substituting our bridge, we get: $$x^2 - 9 \left( \frac{1 - \alpha x}{2} \right)^2 = 9$$ Expanding this, we encounter the term $\frac{1 + \alpha^2 x^2 - 2\alpha x}{4}$. Multiplying the entire equation by $4$ to clear the denominator, we arrive at a structured quadratic equation: $$(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$$ ### The Power of the Discriminant This quadratic is the heart of our problem. It represents the intersection points. If we want the line to never touch the hyperbola, we need this equation to have **no real roots**. In the realm of quadratics, this is the domain of the discriminant, $D = B^2 - 4AC$. We demand that $D < 0$. Calculating $D$ for our equation, where $A = (4 - 9\alpha^2)$, $B = 18\alpha$, and $C = -45$, we find: $$D = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) < 0$$ As we expand this, watch the terms carefully. We get $324\alpha^2 + 180(4 - 9\alpha^2) < 0$. Distributing the $180$, we see $324\alpha^2 + 720 - 1620\alpha^2 < 0$. The terms involving $\alpha^2$ combine to give us $-1296\alpha^2 + 720 < 0$. ### The Final Revelation We are almost there. Rearranging the inequality, we find $1296\alpha^2 > 720$, which simplifies to: $$\alpha^2 > \frac{720}{1296} = \frac{5}{9}$$ Taking the square root, we conclude that **$|\alpha| > \frac{\sqrt{5}}{3}$**. Calculating the decimal value, $\frac{\sqrt{5}}{3} \approx \frac{2.236}{3} \approx 0.745$. This is the threshold of our dance. Any $\alpha$ with an absolute value greater than $0.745$ will ensure the line misses the hyperbola entirely. Looking at our options—$0.5, 0.6, 0.7, 0.8$—only **$0.8$** stands tall above this threshold.

Animated Solution for Mathematics - Conic Sections: If the line $α x + 2 y = 1$ , where $α \in R$ , does not meet the hyperbola $x^{2} - 9 y^{2} = 9$ , then a possible value of $α$ is:

Select Answer:

Visualized Solution

Visualizing the Problem

Standard Form of Hyperbola

Expressing $y$ from the Line

Substitution into Hyperbola

Expanding the Square

Clearing the Denominator

Forming the Quadratic Equation

Condition for No Intersection

Calculating the Discriminant

Simplifying the Inequality

Solving for $α^{2}$

Final Range and Options

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Algebraic Bridge

The Power of the Discriminant

The Final Revelation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line αx+2y=1, where α∈R, does not meet the hyperbola x2−9y2=9, then a possible value of α is: