Conic Sections - Tangents, Normals, Locus Problems
Mathematics 

Conic Sections – Tangents, Normals, Locus Problems

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Introduction

Conic sections—ellipses, hyperbolas, and parabolas—play a crucial role in coordinate geometry and have significant applications in physics, engineering, and astronomy. Understanding their fundamental properties, especially tangents, normals, and locus problems, is essential for solving advanced mathematical problems, including those in competitive exams like IIT JEE.

In this blog, we will explore the equations of tangents and normals to conic sections, deriving key formulas for each type of curve. We will also solve locus problems step-by-step, demonstrating their practical applications. To reinforce these concepts, we’ll analyze previous IIT JEE questions and their solutions, helping students build a strong foundation for tackling similar problems in exams.

Let’s dive into the world of conic sections and uncover their fascinating geometric properties!

I. Tangents to Conic Sections

1.Ellipse

For the ellipse with equation:

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

the equation of the tangent at a point (x_1, y_1) on the ellipse is:

    \[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]

2. Hyperbola

For the hyperbola with equation:

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

the equation of the tangent at a point (x_1, y_1) on the hyperbola is:

    \[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]

3.Parabola

For the parabola with equation:

    \[ y^2 = 4ax \]

the equation of the tangent at a point (x_1, y_1) on the parabola is:

    \[ yy_1 = 2a(x + x_1) \]

II. Normals to Conic Sections

i. Ellipse

For the ellipse, the equation of the normal at (x_1, y_1) is:

    \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 + \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \]

ii. Hyperbola

For the hyperbola, the equation of the normal at (x_1, y_1) is:

    \[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 + \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} \]

iii. Parabola

For the parabola, the equation of the normal at (x_1, y_1) is:

    \[ y - y_1 = -\frac{a}{y_1}(x - x_1) \]

III. Locus Problems

Example Problem 1: Locus of a Point on a Circle

Solution 1:

Let the point P(x, y) be equidistant from A(1, 2) and B(3, 4). The distance from P to A is equal to the distance from P to B, which gives the equation:

    \[ \sqrt{(x - 1)^2 + (y - 2)^2} = \sqrt{(x - 3)^2 + (y - 4)^2} \]

Squaring both sides and simplifying:

    \[ (x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 4)^2 \]

Expanding both sides:

    \[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16 \]

Simplifying:

    \[ 4x + 4y = 20 \]

Thus, the locus of the point is:

    \[ x + y = 5 \]

Example Problem 2: Locus of a Point on a Parabola

Solution 2:

Let the point on the parabola be P(x_1, y_1). The midpoint of the line segment joining the focus F(1, 0) and P(x_1, y_1) is:

    \[ M\left(\frac{x_1 + 1}{2}, \frac{y_1}{2}\right) \]

Since the point P lies on the parabola, y_1^2 = 4x_1. Let M(x, y) represent the midpoint, so:

    \[ x = \frac{x_1 + 1}{2}, \quad y = \frac{y_1}{2} \]

Thus:

    \[ x_1 = 2x - 1, \quad y_1 = 2y \]

Substituting into the equation y_1^2 = 4x_1:

    \[ (2y)^2 = 4(2x - 1) \]

Simplifying:

    \[ 4y^2 = 8x - 4 \]

Thus, the locus is:

    \[ y^2 = 2x - 1 \]

Previous IIT JEE Questions

Question 1: IIT JEE 2019

Solution:

The equation of the tangent to the ellipse \frac{x^2}{4} + \frac{y^2}{9} = 1 at the point P(2, 3) is:

    \[ \frac{x}{2} + \frac{y}{3} = 1 \]

Question 2: IIT JEE 2020

Solution:

The midpoint of the line segment joining the focus F(1, 0) and a point on the parabola y^2 = 4x is:

    \[ M\left(\frac{x_1 + 1}{2}, \frac{y_1}{2}\right) \]

The equation of the locus is:

    \[ y^2 = 2x - 1 \]

 

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