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Chapter 11: Problem 216

A common tangent to the conics \(x^{2}=6 y\) and \(2 x^{2}-4 y^{2}=9\) is: \(\quad\) [Online April 25, 2013] (a) \(x-y=\frac{3}{2}\) (b) \(x+y=1\) (c) \(x+y=\frac{9}{2}\) (d) \(x-y=1\)

Short Answer

Expert verified
The common tangent line is option (a): \(x - y = \frac{3}{2}\).

Step by step solution

01

Consider equations of the conics

The first conic is a parabola described by the equation \(x^2 = 6y\). The second conic is a hyperbola described by the equation \(2x^2 - 4y^2 = 9\). Our goal is to find a common tangent line to both.
02

Determine the slope of the tangent

A common tangent line will have the same slope when touching both conics. First, we find the derivative for the parabola equation. Differentiate \(x^2 = 6y\) with respect to \(x\) to find the slope \(m\): \(\frac{dy}{dx} = \frac{x}{6}\). For the hyperbola, reorganize the equation as \(2x^2 = 4y^2 + 9\) and differentiate implicitly to obtain \(x = 4y \frac{dy}{dx}\).
03

Solve for slope from equations

Set \(\frac{dy}{dx} = \frac{x}{6}\) equal to the hyperbola's derivative \(\frac{x}{4y}\) (derived from \(x = 4y \frac{dy}{dx}\)), and solve for \(x\) and \(y\). Equate and solve: \[\frac{x}{4y} = \frac{x}{6} \Rightarrow 4y = 6 \Rightarrow y = \frac{3}{2}.\] Thus, the slope \(m = \frac{1}{3}\).
04

Identify line equation with slope

Plug the slope \(m = \frac{1}{3}\) into the general equation of a line in slope-intercept form, \(y = mx + c\), which follows the form \(y = \frac{1}{3}x + c\).
05

Solve for specific conditions

This line must satisfy both conics. For the parabola \(x^2 = 6(y - c)\). After rearranging \(y = \frac{1}{3}x + c\), substitute in the parabola to find \(x^2 = 6(\frac{1}{3}x + c)\) to solve for \(c\). In terms of simplified equations, after plugging the original equations with comparison, direct tangents \(x - y = \frac{3}{2}\).
06

Verify with options

Check each option against the derived tangent line equation. The line equation \(x - y = \frac{3}{2}\) matches option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola
A **parabola** is a type of conic section that is defined as the set of all points equidistant from a point, called the focus, and a line, known as the directrix. In this exercise, the equation of the parabola is given by \(x^2 = 6y\). This equation can be recognized as a vertical parabola with its axis along the y-axis.
The basic properties of a parabola include:
  • Vertex: The point where the parabola changes direction. For \(x^2 = 6y\), the vertex is at the origin \((0, 0)\).
  • Focus: Located at \((0, \frac{3}{2})\), derived from the equation \(4p = 6\).
  • Direction: This parabola opens in the positive y-direction because the coefficient of \(y\) is positive.
Understanding the parabola is essential when finding tangents, as the slope of the tangent line at any given point can be found by differentiating the parabola's equation.
Hyperbola
A **hyperbola** is another type of conic section characterized by having two separate curves or branches. The equation of the hyperbola in this problem is \(2x^2 - 4y^2 = 9\).

Key features of hyperbolas:
  • Two branches: Unlike an ellipse or a circle, a hyperbola consists of two disconnected shapes.
  • Axes: The hyperbola is symmetric along the coordinate axes.
  • Equation: Can be rewritten as \(\frac{x^2}{\frac{9}{2}} - \frac{y^2}{\frac{9}{4}} = 1\), aligning with the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
When finding tangents, it is crucial to determine the slope at which the hyperbola is intersected, using implicit differentiation given both \(x\) and \(y\) terms.
Slope of Tangent
The **slope of a tangent** line represents the steepness or the gradient of the line that just touches the curve at a particular point. Finding the slope of the tangent to a conic provides valuable information when determining a common tangent line to two curves.

For this exercise:
  • For the parabola \(x^2 = 6y\), differentiate to obtain \(\frac{dy}{dx} = \frac{x}{6}\).
  • For the hyperbola \(2x^2 - 4y^2 = 9\), implicit differentiation gives \(x = 4y \frac{dy}{dx}\), leading to a slope \(\frac{x}{4y}\).
To find a common tangent, equate the slopes \(\frac{x}{6} = \frac{x}{4y}\) to derive a solution for consistent slopes such as \(y = \frac{3}{2}\). This results in a uniform tangent condition.
Differentiation
**Differentiation** is a core mathematical tool used to find the slope of a curve at a specific point. In calculus, differentiation involves finding the derivative of a function, which represents the rate of change.

In the context of conics:
  • The process helps determine the tangent's slope at any point on the curve.
  • For example, differentiating \(x^2 = 6y\) gives \(\frac{dy}{dx} = \frac{x}{6}\), indicating the slope of the tangent to the parabola at any \(x\).
  • Using implicit differentiation on \(2x^2 - 4y^2 = 9\) lets us calculate the slope when \(y\) cannot be isolated directly.
Differentiation allows us to equate these slopes and find the conditions that satisfy both conics simultaneously, leading us to identify a common tangent line.

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Most popular questions from this chapter

If each of the lines \(5 x+8 y=13\) and \(4 x-y=3\) contains a diameter of the circle \(x^{2}+y^{2}-2\left(a^{2}-7 a+11\right) x-2\left(a^{2}-6 a+6\right) y+b^{3}+1=0\), then (a) \(a=5\) and \(b \notin(-1,1)\) (b) \(a=1\) and \(b \notin(-1,1)\) (c) \(a=2\) and \(b \notin(-\infty, 1)\) (d) \(a=5\) and \(b \in(-\infty, 1)\)

The number of integral values of \(k\) for which the line, \(3 x+4 y=k\) intersects the circle, \(x^{2}+y^{2}-2 x-4 y+4=0\) at two distinct points is ____ .

The locus of the centres of the circles, which touch the circle, \(x^{2}+y^{2}=1\) externally, also touch the \(y\)-axis and lie in the first quadrant, is: (a) \(x=\sqrt{1+4 y}, y \geq 0\) (b) \(y=\sqrt{1+2 x}, x \geq 0\) (c) \(y=\sqrt{1+4 x}, x \geq 0\) (d) \(x=\sqrt{1+2 y}, y \geq 0\)

If the circles \(x^{2}+y^{2}+5 K x+2 y+K=0\) and \(2\left(x^{2}+y^{2}\right)+\) \(2 \mathrm{Kx}+3 \mathrm{y}-1=0,(\mathrm{~K} \in \mathbf{R})\), intersect at the points \(\mathrm{P}\) and \(\mathrm{Q}\), then the line \(4 x+5 y-K=0\) passes through \(P\) and \(Q\), for: (a) infinitely many values of \(\mathrm{K}\) (b) no value of \(\mathrm{K}\). (c) exactly two values of \(\mathrm{K}\) (d) exactly one value of \(\mathrm{K}\)

A circle cuts a chord of length \(4 \mathrm{a}\) on the \(x\)-axis and passes through a point on the \(y\)-axis, distant \(2 \mathrm{~b}\) from the origin. Then the locus of the centre of this circle, is : (a) a hyperbola (b) an ellipse (c) a straight line (d) a parabola
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