Problem 150 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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) \(4 x-2 y=1\)

Step-by-Step Solution

Answer

The equation of the normal is \( 4x - 2y = 1 \), option (d).

Step 1: 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, \( c = 2 \times \frac{1}{2} = 1 \).

Step 2: 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 \( b = \sqrt{3} \).

Step 3: Obtain Equation of the Ellipse

From the calculated values, the standard form of the equation of the ellipse becomes \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \).

Step 4: 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, \( \frac{dy}{dx} = -\frac{3x}{2y} \).

Step 5: 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: \( m_{\text{normal}} = 1 \).

Step 6: 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 \( 4x - 2y = 1 \).

Step 7: Identify Correct Answer Option

The equation of the normal we've determined is \( 4x - 2y = 1 \), hence the correct answer is option (d) \( 4x - 2y = 1 \).

Key Concepts

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.

In the problem, the eccentricity is \(\frac{1}{2}\), meaning the ellipse is not a perfect circle, but also not extremely stretched along its major axis. This value is crucial for understanding the geometry and properties of the ellipse in question. It also helps to determine other parameters such as the semi-major and semi-minor axes.

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.

Understanding the role of the directrix is key because it is related to maintaining the ellipse's shape and its geometric properties. The parameters of the ellipse, including its size and position, are directly influenced by its location and the corresponding eccentricity.

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 \(1\).

This method allows us to calculate the precise equation of the normal line. For the point \(P(1, \beta)\) on the ellipse in the exercise, the normal has been calculated as having the equation \(4x - 2y = 1\). Understanding normals is vital in calculus and geometry, as it helps describe how curves change at various points. It can also have practical applications in physics and engineering, especially when analyzing forces and motions.