Problem 101 - A Complete Resource Book in Mathematics for JEE Main

Question

The locus of the foot of the perpendicular drawn from the centre of the ellipse \(x^{2}+3 y^{2}=6\) 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 correct answer is (B) \((x^2 - y^2)^2 = 6x^2 - 2y^2\).

Step 1: Identify the center and equation of ellipse

Given the ellipse's equation is \(x^2 + 3y^2 = 6\). Compare it to standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) to get \(a^2 = 6\) and \(b^2 = 2\). The center of the ellipse is \((0, 0)\).

Step 2: Equation of tangent to ellipse

The general form for a tangent to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\). Substitute \(a^2 = 6\) and \(b^2 = 2\) to get the tangent equation \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\).

Step 3: Find the foot of the perpendicular

For any point \((x_1, y_1)\) on this ellipse, use the tangent equation \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\). The equation of the line passing through \((0,0)\) perpendicular to this tangent is \(\frac{x}{x_1} = \frac{y}{y_1}\).

Step 4: Combine the tangent and perpendicular equations

Substitute \(y = \frac{yx_1}{x} \) from the perpendicular equation into \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\) to find the locus. Simplifying, we arrive at the locus as \((x^2 - y^2)^2 = 6x^2 - 2y^2\).

Step 5: Select the correct answer

Compare the derived equation of the locus \((x^2 - y^2)^2 = 6x^2 - 2y^2\) with the options provided. The correct answer corresponds to option (B).

Key Concepts

Locus of a point

The term "locus" in mathematics refers to a set of points that satisfy a specific condition or a set of conditions. Imagine tracing a path where each point falls under a particular rule. In the context of ellipses, a locus might describe certain geometric properties, like the path traced by the foot of a perpendicular from some significant point, such as the center of the ellipse.

In the exercise, we're examining the locus of the foot of the perpendicular drawn from the center of the ellipse to any tangent line. To put it in simple terms, we are interested in the path or curve formed by these foot points as the tangent moves around the ellipse.

An easy way to visualize this is to imagine the ellipse as a deformed circle, where the locus of points often takes on unique shapes that highlight the symmetry and properties of the ellipse. Understanding loci in this scenario can provide deeper insights into the nature of ellipses and their tangent lines.

Tangent to an ellipse

A tangent line to an ellipse is a straight line that just touches the ellipse at one point, without crossing it. The equation of a tangent line can be determined by employing the general form for a tangent of an ellipse.

For the ellipse given by the equation \(x^2 + 3y^2 = 6\), we convert it into the standard form: \(\frac{x^2}{6} + \frac{y^2}{2} = 1\). From here, the equation of a tangent line at any point \((x_1, y_1)\) on the ellipse can be expressed as:\

• \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\).

This formula arises from a special relationship in ellipses that tangents at any point \((x_1, y_1)\) on the ellipse can be written in this form.

Understanding the tangent here is crucial because it provides the necessary relationship to find the foot of the perpendicular from the center of the ellipse, which helps determine the path or locus of points we are searching for.

Perpendicular from a point to a line

A perpendicular line segment from a point to a line is the shortest possible distance between that point and the line. This segment intersects the line at a right angle (90 degrees).

In this problem, we consider a perpendicular drawn from the center of an ellipse, represented by the point (0,0), to a tangent line. The line equation that is perpendicular to the tangent and passes through the center of the ellipse can be expressed as

• \(\frac{x}{x_1} = \frac{y}{y_1}\),

where \((x_1, y_1)\) lies on the ellipse.

This perpendicular relationship is essential as it defines the position of the foot of the perpendicular. The foot is the point of intersection between this perpendicular line and the tangent. By determining the unique foot point for every tangent, we are essentially mapping out the locus of these foot points, unlocking the complete geometric picture of how these constraints define specific paths or curves for an ellipse.