Question

A common tangent to \(9 \mathrm{x}^{2}-16 \mathrm{y}^{2}=144\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\) is (a) \(\mathrm{y}=3 \sqrt{(2 / 7) \mathrm{x}+(15 / \sqrt{7})}\) (b) \(\mathrm{y}=2 \sqrt{(3 / 7) \mathrm{x}-15 \sqrt{7}}\) (c) \(\mathrm{y}=(3 / \sqrt{7}) \mathrm{x}+(15 / \sqrt{7})\) (d) \(\mathrm{y}=2 \sqrt{(3 / 7) \mathrm{x}+15 \sqrt{7}}\)

Short answer

The common tangent to the given curves is: \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\)

Step-by-step solution

(Step 1: Rewrite the equations in standard form)

Divide the equation of the hyperbola by 144 to get the standard form: 1) \(\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1\) and the equation for circle is already in standard form: 2) \(x^2 + y^2 = 9\)

(Step 2: Find the tangent of the hyperbola using parameter t)

The equation of the tangent of the hyperbola in the parametric form is : \(y = tx + \frac{3}{4t}\)

(Step 3: Check the same tangent on the circle, by substituting the point and slope from the tangent equation of a hyperbola)

Substitute the point on the hyperbola tangent: x = 4\(\sqrt{\frac{1 + t^2}{t^2}}\); y = 3\(\sqrt{\frac{1 + t^2}{1 - t^2}}\) Now rewrite the tangent equation as : \(y = tx - 3t^3 + 3t\)

(Step 4: Compare the resulting equation with the given options to get the correct answer)

Comparing the resulting equation with the given options, we find that the correct answer is: (b) \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\) Hence, the common tangent to the given curves is: \(y = 2\sqrt{\frac{3}{7}x - 15\sqrt{7}}\)

Key concepts

Hyperbola

A hyperbola is a type of conic section that is formed by intersecting a double cone. In algebraic terms, it is represented by a quadratic equation in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). This standard form helps us identify important properties like the vertices and asymptotes.

• Vertices are located at \((\pm a, 0)\) or \((0, \pm b)\), depending on the orientation of the hyperbola.
• Asymptotes, the lines that the curve approaches but never actually meets, help in sketching the curve accurately. For the given hyperbola \(9x^2 - 16y^2 = 144\), when rewritten, becomes \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \).

This shows it is centered at the origin, \((0, 0)\), with focal points located along the x-axis. Hyperbolas are unique compared to other conic sections due to their open curve nature and two separate branches.

Circle

In contrast to a hyperbola, a circle is a simple closed shape where every point on its edge is equidistant from a central point. The standard form of its equation is \( x^2 + y^2 = r^2 \). Here, \(r\) stands for the radius of the circle.
For the given problem, the circle equation \(x^2 + y^2 = 9\) tells us:

• The circle is centered at the origin, \((0, 0)\).
• Its radius is \(3\), as \(r^2 = 9\).

Circles are crucial in geometry due to their symmetrical properties. Visualizing a circle can help in understanding relationships with other curves, such as their intersections and tangents, as it is symmetrical in every direction from its center.

Parametric Form

The parametric form is a way of expressing equations whereby the coordinates of the points are defined using a parameter, often denoted as \(t\). This form is extensively used in curves like circles and hyperbolas because it allows for easy manipulation of the curves' equations.
For the hyperbola \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \), the parametric equations are provided by:

• \( x = 4 \sqrt{1 + t^2} \)
• \( y = 3 \sqrt{1 + t^2}/t \)

Such expressions help us explore the paths traced by points along the curve as \(t\) varies. This form is also valuable in deriving tangents and normals to a curve since it smooths out computations linked to varying slope scenarios.

Tangent Equation

A tangent to a curve at a point is a straight line that just "touches" the curve at that point. For mathematical understanding, it does not cross over or remain on the curve for any appreciable distance. Rather, it represents the slope of the curve at that particular point.
To find a tangent's equation in any case, identifying a point on the curve and determining the slope, often denoted as \(m\), is crucial. With these two, the tangent formula becomes:

• In parametric terms: \( y = tx + \frac{3}{4t} \)
• This is for the hyperbola's tangent.

Finding tangents to curves like hyperbolas involves using derivatives or other algebraic methods. This concept is significant because tangents can often represent real-world constraints, limitations, or interactions of varying geometric settings.