Problem 101 - A Complete Resource Book in Mathematics for JEE Main 2017

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: Write the equation of the ellipse

The given ellipse is \(x^2 + 3y^2 = 6\). By rearranging, we can obtain the standard form: \(\frac{x^2}{6} + \frac{y^2}{2} = 1\). This shows that the center of the ellipse is at the origin \((0,0)\).

Step 2: Write the equation of the tangent

The equation of the tangent to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) in the point-slope form is \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\). For our ellipse, this becomes \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\) for a point \((x_1, y_1)\) on the ellipse.

Step 3: Find the foot of the perpendicular

Consider that the center of the ellipse \((0,0)\) is at a perpendicular distance from the tangent. Plug \((0, 0)\) into the equation of the tangent: \(\frac{0 \cdot x_1}{6} + \frac{0 \cdot y_1}{2} = 1\). This implies the perpendicular passes through \((x_{0}, y_{0})\), simplifying to \(x_0x_1 + 3y_0y_1 = 6\).

Step 4: Square both sides and simplify

To find the locus of \((x_0, y_0)\), replace \(x_1\) and \(y_1\) using the parametric equations \(x_1 = a \cos \theta\), \(y_1 = b \sin \theta\). Substitute these back into the equation: \(a^2 \cos^2 \theta = 6 - 6 \sin^2 \theta\) and \(b^2 \sin^2 \theta = 3(2 - 2 \cos^2 \theta)\). Squaring and substituting gives \((x_0^2 - y_0^2)^2 = 6x_0^2 -2y_0^2\) upon simplification.

Key Concepts

ellipse

An ellipse is a fascinating geometric shape that looks like a stretched-out circle. In mathematical terms, it's defined as the set of all points where the sum of the distances from two fixed points, called the foci, is constant. The general equation of an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

• The semi-major axis is the longest diameter of the ellipse.
• The semi-minor axis is the shortest diameter.

For the specific ellipse given in the exercise, \(x^2 + 3y^2 = 6\), you can see that the center is at the origin \((0,0)\) and it is rearranged into standard form as \(\frac{x^2}{6} + \frac{y^2}{2} = 1\) to determine the axes: \(a^2 = 6\) and \(b^2 = 2\). Ellipses are fundamental in fields like astronomy, where they describe the orbits of planets.

perpendicular

A perpendicular line is one that makes a right angle, or 90 degrees, with another line. In the context of this exercise, we need to find the line perpendicular to a tangent of the ellipse, starting from its center. This is crucial because the exercise involves finding the locus, or path, traced by such perpendiculars.

• Perpendicular lines are a key concept in geometry because they are used to create right angles.
• In problems involving perpendiculars, the slope or gradient condition is often used: If two lines are perpendicular, the product of their slopes is \(-1\).

In this exercise, the concept of perpendicular distance is used to understand how the center of the ellipse relates spatially to the tangent.

tangent line

A tangent line is one that touches a curve at only one point. At that point, it is parallel to the curve's direction. For an ellipse like \(\frac{x^2}{6} + \frac{y^2}{2} = 1\), the equation of a tangent line at any point \((x_1, y_1)\) is \(\frac{xx_1}{6} + \frac{yy_1}{2} = 1\).

• Understanding tangent lines is essential in calculus and geometry because they describe how curves behave at points.
• Tangents provide insights into the slope and direction of a curve, playing a key role in defining smooth paths.

For this problem, understanding how to derive and use the tangent line equation is crucial in finding the location where the perpendicular meets it from the ellipse's center.

foot of perpendicular

The foot of the perpendicular is the point where a perpendicular line from a specific point meets another line. In the context of the ellipse problem, it relates to where the perpendicular from the center of the ellipse \((0,0)\) intersects the tangent line to the ellipse.

• It is essentially the 'landing point' of the perpendicular on the other line.
• Identifying this helps in determining the path, or locus, of points that describe certain geometric properties.

In this exercise, finding the foot of the perpendicular helps uncover the set of points (locus) where such perpendiculars drawn from the center of the ellipse meet various tangents. The locus equation \((x_0^2 - y_0^2)^2 = 6x_0^2 - 2y_0^2\) ultimately helps represent this set of points.