Problem 179 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Equation of the line passing through the points of intersection of the parabola \(x^{2}=8 y\) and the ellipse \(\frac{x^{2}}{3}+y^{2}=1\) is: \(\quad\) [Online April 9, 2013] (a) \(y-3=0\) (b) \(y+3=0\) (c) \(3 y+1=0\) (d) \(3 y-1=0\)

Step-by-Step Solution

Answer

The equation of the line is \(3y - 1 = 0\).

Step 1: Equate the Curves

First, we seek the points of intersection by setting the parabola and ellipse equal to one another. From the parabola, solve for \( y \): \( y = \frac{x^2}{8} \). Substitute into the ellipse equation: \[ \frac{x^2}{3} + \left( \frac{x^2}{8} \right)^2 = 1. \]

Step 2: Simplify the Equation

Simplify the substitution into the ellipse equation: \[ \frac{x^2}{3} + \frac{x^4}{64} = 1. \] Multiply through by 64 to eliminate the fractions: \[ \frac{64x^2}{3} + x^4 = 64. \] This simplifies to \[ 3x^4 + 64x^2 - 192 = 0. \]

Step 3: Solve for x

Factor the equation \( 3x^4 + 64x^2 - 192 = 0 \) by setting \( x^2 = t \), obtaining \( 3t^2 + 64t - 192 = 0 \). Solve for \( t \) using the quadratic formula: \[ t = \frac{-64 \pm \sqrt{64^2 + 4 \cdot 3 \cdot 192}}{2 \cdot 3} = \frac{-64 \pm 80}{6}. \] This gives \( t = \frac{16}{6} = \frac{8}{3} \) and \( t = -24 \). Since \( t = x^2 \), we discard \( t = -24 \).

Step 4: Find the y-values

Back-substitute \( t = \frac{8}{3} \) to find \( x \): \( x^2 = \frac{8}{3} \). Therefore, \( y = \frac{x^2}{8} = \frac{1}{3} \). Thus, the point of intersection is \( (\sqrt{\frac{8}{3}}, \frac{1}{3}) \) and \( (-\sqrt{\frac{8}{3}}, \frac{1}{3}) \) due to symmetry.

Step 5: Equation of the Line

For a horizontal line (since both y-values are identical), the line equation is \( y = \frac{1}{3} \). Comparing it to the options, rewrite: \( 3y = 1 \), or rearrange to: \( 3y - 1 = 0. \)

Key Concepts

Parabola

A parabola is a unique and interesting curve with a characteristic U-shape. It can be expressed in an equation of the form

• \( y = ax^2 + bx + c \)

where the cofficient \(a\) determines the direction and stiffness of the parabola's open sides. In coordinate geometry, the focus is on understanding its properties.

For the given problem, the parabola represented is \( x^2 = 8y \), which is a special type of parabola called a "vertical parabola" because it opens upward. Here:

• The vertex is at the origin \((0, 0)\).
• The parabola extends infinitely up or down depending on the coefficient, but always maintains its symmetrical shape.
• The axis of symmetry is the vertical line passing through the vertex, in this case, the y-axis.

To find the intersection with another curve, we solve simultaneously, as was done with the ellipse in the solution.

Ellipse

An ellipse is like a stretched circle and is defined as the collection of all points for which the sum of the distances to two fixed points (foci) is constant. Its general equation is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Here, \( a \) and \( b \) are the semi-major and semi-minor axes respectively. The ellipse in the problem is expressed as \( \frac{x^2}{3} + y^2 = 1 \), showing it is centered at the origin \((0,0)\).

• This ellipse is horizontally oriented (wider along the x-axis), because \( a = \sqrt{3} \) is greater than \( b = 1 \).
• Every ellipse has two focal points and is perfectly symmetrical.

For the problem at hand, we substituted the expression obtained from the parabola into the ellipse equation to find the points of intersection.

These points are where the parabola and ellipse's paths cross, indicating where both equations are true at the same time.

Equation of a line

The equation of a line in its most common form is given by:

\( y = mx + c \),

where \( m \) is the line's slope and \( c \) is the y-intercept. A unique property about lines is that if two points have the same y-value, as in our exercise, the line passing through these points is horizontal. When there is no slope (\( m = 0 \)), the equation becomes simply \( y = c \).

• For the horizontal line in the solution, we derived the equation \( y = \frac{1}{3} \) by recognizing that the points from the intersection had the same y-value (\( \frac{1}{3} \)).
• We then expressed it as \( 3y - 1 = 0 \), which is one of the given choices.

This process shows how using intersection points, even from different shapes, boils down to recognizing and using consistent y-values to denote a straight line through these points.