Problem 173 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The locus of the foot of perpendicular drawn from the centre of theedlipse \(x^{2}+3 y^{2}=6\) on any tangent to it is (a) \(\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}\) (b) \(\left(x^{2}+y^{2}\right)^{2}=6 x^{2}-2 y^{2}\) (c) \(\left(x^{2}-y^{2}\right)^{2}=6 x^{2}+2 y^{2}\) (d) \(\left(x^{2}-y^{2}\right)^{2}=6 x^{2}-2 y^{2}\)

Step-by-Step Solution

Answer

The answer is (b): \((x^2 + y^2)^2 = 6x^2 - 2y^2\).

Step 1: Identify Given Equation

The given ellipse is \( x^2 + 3y^2 = 6 \). This can be rewritten as \( \frac{x^2}{6} + \frac{y^2}{2} = 1 \).

Step 2: Standard Form and Center

The equation \( \frac{x^2}{6} + \frac{y^2}{2} = 1 \) is in the standard form of an ellipse, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The center of this ellipse is at the origin, (0,0).

Step 3: Equation of the Tangent

The equation of a tangent to the ellipse at point \((x_1, y_1)\) is \( \frac{xx_1}{6} + \frac{yy_1}{2} = 1 \).

Step 4: Foot of Perpendicular

Let's consider the point \((h, k)\) as the foot of the perpendicular from the center \((0,0)\) to the tangent. The perpendicular from \((0,0)\) meets the tangent at \((h, k)\) along the line through the origin such that the vector \((h, k)\) is normal to the tangent line vector \((\frac{x_1}{6}, \frac{y_1}{2})\).

Step 5: Derive the Condition

Since \((h, k)\) lies on the tangent, substituting into the tangent equation, we have \( \frac{hh}{6} + \frac{kk}{2} = 1 \), or \( \frac{h^2}{6} + \frac{k^2}{2} = 1 \).

Step 6: Forming the Locus Equation

By manipulating and eliminating parameters, squaring and adding \(h^2\) and \(k^2\) equations leads to identifying the locus of \((h, k)\) as: \( \frac{h^2}{6} + \frac{k^2}{2} = 1 \). Multiplying throughout by 12 to simplify, the equation becomes \( 2h^2 + 6k^2 = 12 \), further reducing and manipulating to form \( (x^2 + y^2)^2 = 6x^2 - 2y^2 \).

Step 7: Conclusion

After examining the derived equation in step 6, the correct choice for the locus is option (b): \( (x^2 + y^2)^2 = 6x^2 - 2y^2 \).

Key Concepts

Ellipse Equation

An ellipse is a fundamental geometric shape that appears in various fields such as physics, engineering, and astronomy. The general form of an ellipse equation is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively.
In the context of this exercise, the equation \( x^2 + 3y^2 = 6 \) is rewritten in standard form as \( \frac{x^2}{6} + \frac{y^2}{2} = 1 \). Here, \( 6 \) and \( 2 \) correspond to \( a^2 \) and \( b^2 \), outlining the proportions of the ellipse along the x and y axis.

• The center of the ellipse is at the origin (0,0).
• The axis lengths determine the shape and orientation of the ellipse.

This formulation is essential as it helps in understanding how the ellipse interacts with lines and shapes drawn in relation to it, like tangents in our case.

Tangent to Ellipse

A tangent to an ellipse is a straight line that touches the ellipse at exactly one point. This point is called the point of tangency. For any point \((x_1, y_1)\) on the ellipse, the equation of the tangent line can be found using the formula \( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \).
From the specific ellipse equation \( x^2 + 3y^2 = 6 \), the tangent line at a point can be expressed as \( \frac{xx_1}{6} + \frac{yy_1}{2} = 1 \).

• The tangent line is perpendicular to the radius vector, which is the line drawn from the center of the ellipse to the point of tangency.
• This property is crucial when finding perpendiculars from the center of the ellipse to any tangent.

Understanding how to formulate and visualize tangents helps in analyzing movement and distances in ellipse geometry, leading toward solving locus problems.

Perpendicular from Center

The perpendicular from the center of the ellipse to a tangent line plays an integral role in finding certain geometric loci. The center of the ellipse, in our case, is \((0,0)\).
To find the foot of the perpendicular from the center to a tangent, we rely on the normalcy between the radius (from the center to the foot) and the tangent line.

• Given the normal vector to the tangent \( \left( \frac{x_1}{6}, \frac{y_1}{2} \right) \), the perpendicular line from the center would also respect this vector's direction.
• The point \((h, k)\) where the perpendicular intersects the tangent, satisfies the tangent equation.

The calculation of such a point \((h, k)\) forms the basis for determining more complex coordinates like the locus of points relative to an ellipse.

Ellipse Geometry

Ellipse geometry is the study of properties and relationships within ellipses, focusing on aspects like foci, axes, and intersecting lines.
One interesting topic is the locus of points, such as the foot of perpendiculars from the center of an ellipse to various tangents. These locus points create specific curves or paths. In the exercise, the task is to find the locus of the foot of the perpendiculars, requiring an understanding of how these points relate geometrically to the defined ellipse.

• By manipulating the tangent equation and conditions, we derive an expression representing the locus.
• The locus here simplifies to an equation \( (x^2 + y^2)^2 = 6x^2 - 2y^2 \), which describes the set path of all such perpendiculars' intersections.

Recognizing these geometric relationships allows us to derive equations that describe intricate motion and positional attributes within an elliptical space.