Problem 34 - A Complete Resource Book in Mathematics for JEE Main

Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse \(x^{2}\) \(+2 y^{2}=2\) between the coordinate axes is (A) \(\frac{1}{x^{2}}+\frac{1}{2 y^{2}}=1\) (B) \(\frac{1}{4 x^{2}}+\frac{1}{2 y^{2}}=1\) (C) \(\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1\) (D) \(\frac{1}{2 x^{2}}+\frac{1}{y^{2}}=1\)

Step-by-Step Solution

Answer

The correct answer is (C).

Step 1: Understand the Problem

We are tasked with finding the locus of the midpoint of intercepts made by the tangents from an external point to the given ellipse \(x^2 + 2y^2 = 2\) on the coordinate axes. The key is to determine this midpoint's equation in terms of its coordinates \((h, k)\).

Step 2: Equation of Tangents from an External Point

For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the equation of tangents from \((x_0, y_0)\) is \(\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1\). For the ellipse \(x^2 + 2y^2 = 2\), rewriting in standard form gives \(\frac{x^2}{2} + \frac{y^2}{1} = 1\). So, the tangent's equation is \(\frac{xx_0}{2} + \frac{yy_0}{1} = 1\).

Step 3: Find Intercepts on Coordinate Axes

Setting \(y = 0\) in the tangent equation, we get \(\frac{xx_0}{2} = 1\) leading to \(x = \frac{2}{x_0}\). For \(x = 0\), the tangent equation gives \(yy_0 = 1\), leading to \(y = \frac{1}{y_0}\). So intercepts are \((\frac{2}{x_0}, 0)\) and \((0, \frac{1}{y_0})\).

Step 4: Midpoint of Intercepts

The midpoint of intercepts \((\frac{2}{x_0}, 0)\) and \((0, \frac{1}{y_0})\) is \(\left(\frac{1}{x_0}, \frac{1}{2y_0}\right)\). Let's denote this point as \((h, k)\), so we have \(h = \frac{1}{x_0}\) and \(k = \frac{1}{2y_0}\).

Step 5: Substitute Back for Locus Equation

From Step 4, we found \(x_0 = \frac{1}{h}\) and \(y_0 = \frac{1}{2k}\). So the original equation of the ellipse \(\frac{x_0^2}{2} + \frac{y_0^2}{1} = 1\) becomes \(\frac{(\frac{1}{h})^2}{2} + \frac{(\frac{1}{2k})^2}{1} = 1\). Simplifying gives \(\frac{1}{2h^2} + \frac{1}{4k^2} = 1\).

Step 6: Identify the Correct Option

The simplified equation \(\frac{1}{2h^2} + \frac{1}{4k^2} = 1\) matches with option (C) when replacing \(h\) with \(x\) and \(k\) with \(y\). So the correct answer is (C).

Key Concepts

Ellipse Geometry

An ellipse is a set of points in a plane where the sum of the distances from two fixed points, called foci, is constant. This shape resembles a flattened circle, and its geometry has some unique properties that are important in coordinate geometry.
One of the main features of an ellipse is its symmetry across its two axes. The line through the foci is called the major axis, while the line perpendicular to the major axis and through the center is the minor axis. The lengths of these axes determine the shape of the ellipse. If the major and minor axes are equal, the ellipse becomes a circle.
For an ellipse expressed in standard form, \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] the lengths of the semi-major and semi-minor axes are denoted by \(a\) and \(b\) respectively. A distinct characteristic of an ellipse is that if \(a > b\), the major axis lies along the x-axis and vice versa. Understanding the structure of an ellipse is crucial for determining other related properties, such as tangent lines and the locus of points.

Tangent to Ellipse

A tangent to an ellipse is a line that touches the ellipse at exactly one point. At this point of contact, the tangent does not cross the ellipse, but only touches it gently. The concept of tangents is pivotal in understanding how external points relate to the ellipse.
The equation of a tangent to an ellipse at any point \( (x_0, y_0) \) can be expressed as\[ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1 \] where \(a\) and \(b\) are the semi-major and semi-minor axes of the ellipse, respectively.
To find tangents from an external point, the point from which the tangents are drawn is substituted into the tangent equation, adjusting for the given ellipse's parameters. In coordinate geometry, determining where tangents intersect the axes involves setting \(x\) and \(y\) to zero alternately to find the intercepts. These intercepts assist in calculating various geometric properties, such as the midpoint of the intercepts or the locus of points which forms a new shape or equation.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, utilizes algebraic formulas to study geometric figures. This branch of mathematics provides a powerful toolset for describing and analyzing the spatial relationships and properties of geometric shapes like ellipses.
In this context, the locus of a point is the set of all points satisfying specific conditions. For example, in this exercise, the locus of the midpoint of intercepts formed by tangents from an external point to the ellipse produces a whole new curve or expression. This occurs by analyzing the relationships that hold between intercepts, coordinates, and the ellipse's typical properties.
Coordinate geometry helps to simplify complex geometric problems by converting them into algebraic equations, which can then be manipulated and solved. Using the axes as reference, points can be translated, rotated, and scaled, making this field versatile for a myriad of applications.

• The intersection of lines and curves.
• Identifying loci of points.
• Analyzing symmetry and transformations.

These tools aid mathematicians and engineers alike in designing, modeling, and understanding structures and dynamics within both the natural and engineered world.