Problem 17 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation \(2 x^{2}+3 y^{2}-8 x-18 y+35=k\) represents (A) no locus if \(k>0\) (B) an ellipse if \(k<0\) (C) a point if \(k=0\) (D) a hyperbola if \(k>0\)

Step-by-Step Solution

Answer

The equation represents (A) no locus if \(k > 0\).

Step 1: Understanding the Given Equation

We begin by examining the given equation: \(2x^2 + 3y^2 - 8x - 18y + 35 = k\). Our goal is to determine the nature of the locus represented by this equation based on the value of \(k\).

Step 2: Complete the Square for x

Isolate terms involving \(x\): \(2x^2 - 8x\). Complete the square: \(2(x^2 - 4x)\). Express \(x^2 - 4x\) as \((x-2)^2 - 4\) to get \(2((x-2)^2 - 4) = 2(x-2)^2 - 8\).

Step 3: Complete the Square for y

Separate terms involving \(y\): \(3y^2 - 18y\). Complete the square: \(3(y^2 - 6y)\). Express \(y^2 - 6y\) as \((y-3)^2 - 9\) to have \(3((y-3)^2 - 9) = 3(y-3)^2 - 27\).

Step 4: Rewrite the Equation

Combine these completed squares back into the equation, rewriting it as \(2(x-2)^2 + 3(y-3)^2 - 8 - 27 + 35 = k\). Simplify the constants: \(-8 - 27 + 35 = 0\). So, the equation becomes \(2(x-2)^2 + 3(y-3)^2 = k\).

Step 5: Analyzing the Resulting Equation

Now we determine the nature of the conic: \(2(x-2)^2 + 3(y-3)^2 = k\). **Ellipse case**: If \(k > 0\), this is the equation of an ellipse because both coefficients of the squared terms are positive. **Point case**: If \(k = 0\), the conic reduces to a point at \((2,3)\). **No real locus**: If \(k < 0\), there is no real solution since squared terms cannot sum to a negative value.

Key Concepts

Ellipse

An ellipse is a type of conic section that resembles an elongated circle. It is the set of all points in a plane such that the sum of the distances from two fixed points (called foci) is constant. This results in a smooth, closed curve. Ellipses can appear more stretched than circles, depending on how distant the foci are from each other.

An equation representing an ellipse in its standard form looks like this:

• \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)

Here, \((h, k)\) are the coordinates of the center of the ellipse, while \(a\) and \(b\) denote the semi-major and semi-minor axes, respectively.

In our exercise, the equation transformed into the form \(2(x-2)^2 + 3(y-3)^2 = k\), resembles an ellipse when \(k > 0\) because each term of \(x\) and \(y\) is positive. This ensures the conic section is indeed an ellipse and not another form like a hyperbola. The coefficients \(2\) and \(3\) indicate the stretching along the respective axes.

Hyperbola

A hyperbola is another form of a conic section, distinctively different from ellipses and circles. It consists of two separate curves (branches) that mirror each other. The set of all points that form a hyperbola have the same absolute difference in distances to two fixed points called foci.

The standard equation of a hyperbola is written as:

• \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• or \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \)

These equations denote a horizontal or vertical orientation for the branches of the hyperbola.

However, in the given exercise, the equation \(2(x-2)^2 + 3(y-3)^2 = k\) suggests a hyperbola does not form because both of the squared terms are positive. For a hyperbola to exist from this form of equation, one of these terms should have a negative coefficient.

Completing the Square

Completing the square is a strategy used often in algebra to transform quadratic equations into a form that is much easier to analyze and work with. This method is particularly useful in converting conic section equations into their standard forms, like the example with our exercise.

This process involves rearranging a quadratic expression into a perfect square trinomial. It follows these basic steps:

• Take the quadratic expression, e.g., \(ax^2 + bx\).
• Factor out the leading coefficient from the linear terms if necessary.
• Find \(\left( \frac{b}{2a} \right)^2\), add and subtract this square inside the equation to complete it.
• Rewrite the expression as a perfect square, \((x-d)^2 - (constant)\).

For our exercise, we applied this technique to both the \(x\) and \(y\) terms. This allowed us to rewrite the original equation into the form \(2(x-2)^2 + 3(y-3)^2 = k\), from which students can easily identify the nature of the curve as they change the value of \(k\). Such transformations are critical in studying the characteristics and classification of conic sections.