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Magazine Chapter 11: Problem 150
Let \(x=4\) be a directrix to an ellipse whose centre is at the origin and its eccentricity is \(\frac{1}{2}\). If \(P(1, \beta), \beta>0\) is a point on this ellipse, then the equation of the normal to it at \(P\) is : (a) \(4 x-3 y=2\) (b) \(8 x-2 y=5\) (c) \(7 x-4 y=1\) (d) \(4 x-2 y=1\)
Short Answer
Expert verified
The equation of the normal is \( 4x - 2y = 1 \), option (d).
Step by step solution
01
Determine Relationship Between Parameters
The ellipse's eccentricity is given as \( e = \frac{1}{2} \). The directrix is a vertical line at \( x = 4 \). For an ellipse, the relationship among the semi-major axis \( a \), the semi-minor axis \( b \), and the eccentricity \( e \) is defined by \( e = \frac{c}{a} \) where \( c = \sqrt{a^2 - b^2} \). Since \( x = 4 \) is a directrix, we use the formula \( a/e = 4 \), giving \( a = 2 \). Consequently, \( c = 2 \times \frac{1}{2} = 1 \).
02
Use Relationship to Find \\( b \\)
Now substitute \( a = 2 \) and \( c = 1 \) in the equation \( c^2 = a^2 - b^2 \). Thus, \( 1 = 4 - b^2 \). Solving, we obtain \( b^2 = 3 \) or \( b = \sqrt{3} \).
03
Obtain Equation of the Ellipse
From the calculated values, the standard form of the equation of the ellipse becomes \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \).
04
Differentiate the Ellipse Equation
Using implicit differentiation, differentiate \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \) with respect to \( x \):\[ \frac{2x}{4} + \frac{2y}{3} \cdot \frac{dy}{dx} = 0 \]. Simplifying gives \( \frac{x}{2} + \frac{y}{3} \cdot \frac{dy}{dx} = 0 \). So, \( \frac{dy}{dx} = -\frac{3x}{2y} \).
05
Find Slope of Normal at Point \\( P(1, \beta) \\)
Calculate \( \beta \) from the equation of the ellipse: \( \frac{1^2}{4} + \frac{\beta^2}{3} = 1 \) yielding \( \frac{1}{4} + \frac{\beta^2}{3} = 1 \). Solving, \( \frac{\beta^2}{3} = \frac{3}{4} \) gives \( \beta^2 = \frac{9}{4} \), or \( \beta = \frac{3}{2} \). Then calculate \( \frac{dy}{dx} \) at \( (1, \frac{3}{2}) \), which is \( -\frac{3 \times 1}{2 \times \frac{3}{2}} = -1 \). The slope of the normal is the negative reciprocal: \( m_{\text{normal}} = 1 \).
06
Form Equation of Normal
The equation of the line with slope 1 passing through \( (1, \frac{3}{2}) \) is found using the point-slope formula: \( y - \beta = m_{\text{normal}}(x - 1) \). Substituting, \( y - \frac{3}{2} = 1(x - 1) \). This simplifies to \( y = x - 1 + \frac{3}{2} \) or \( y = x + \frac{1}{2} \), which is equivalent to \( 2y = 2x + 1 \) or \( 2x - y = -1 \). Adding 2 to both sides, the equation becomes \( 4x - 2y = 1 \).
07
Identify Correct Answer Option
The equation of the normal we've determined is \( 4x - 2y = 1 \), hence the correct answer is option (d) \( 4x - 2y = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eccentricity
Eccentricity is a fundamental concept in understanding the shape of an ellipse. Imagine pulling a rubber band around two pins on a board. The more spread apart these pins, the more elongated the shape of the band becomes, simulating the eccentricity of an ellipse.
- Eccentricity ( \(e\) ) of an ellipse measures how much it deviates from being a perfect circle. If \(e = 0\), it is a circle.
- For any ellipse, \(0 < e < 1\). The closer \(e\) is to 1, the more stretched the ellipse appears.
- The relationship is given by \(e = \frac{c}{a}\), where \(c\) is the distance from the center to each focus, and \(a\) is the semi-major axis.
Directrix
A directrix is a handy tool when dealing with ellipses. It's a line used in conjunction with a focus to define an ellipse geometrically. Imagine drawing an ellipse using a thumbtack (focus) and a pencil with a string tied to a straight edge (directrix).
- For an ellipse, the directrix is a line that, together with a focus, helps in maintaining a constant ratio, which is the eccentricity.
- In mathematical terms, for any point \(P\) on the ellipse, the ratio of its distance to a focus (let's call it \(F\)) and its distance to the directrix is the eccentricity \(e\): \(\frac{PF}{PD} = e\).
- In the provided exercise, the line \(x = 4\) served as the directrix, which means it guides the shape of the ellipse alongside its focus.
Normal to Ellipse
The concept of a normal to an ellipse is essential in understanding the tangent and perpendicular lines at any point on the ellipse. Imagine drawing a line perpendicular to the surface at a point on an egg. This line represents the normal.
- A normal to an ellipse at any point is a line that is perpendicular to the tangent at that point.
- In the exercise, finding the normal required first determining the slope of the tangent using implicit differentiation.
- The slope of the normal is the negative reciprocal of the tangent slope. For instance, if the slope of the tangent is \(-1\), then the normal has a slope of \(1\).
