The equation 2x² + 3y² - 8x - 18y + 35 = k represents (a) no locus if k 0 (b) an ellipse if k

Q: Wait, why do we reject the negative root or the case where $k < 0$?

Great question! Look at our final simplified equation: $2(x - 2)^2 + 3(y - 3)^2 = k$. Since $(x - 2)^2$ and $(y - 3)^2$ are squares of real numbers, they must be non-negative. If $k < 0$, we are essentially saying that the sum of two non-negative numbers is negative, which is impossible in the real number system. Thus, there is no real locus.

Q: Why can't we just say it's an ellipse immediately because of the $x^2$ and $y^2$ terms?

It is tempting to jump to conclusions! While the coefficients of $x^2$ and $y^2$ being positive and unequal is a strong indicator of an ellipse, it is not the whole story. A degenerate conic, like a single point, also fits this general form. We must complete the square to see if the constant on the right side allows for a real, extended shape or collapses into a single point.

Q: Is it possible for this to be a circle?

For a conic section to be a circle, the coefficients of $x^2$ and $y^2$ must be identical. In our equation, we have $2x^2$ and $3y^2$. Because $2 \neq 3$, the stretching factor in the $x$-direction is different from the $y$-direction, which is the definition of an ellipse, not a circle.

Question

The equation $2x^2 + 3y^2 - 8x - 18y + 35 = k$ represents

Options:

(A) no locus if $k > 0$

(B) an ellipse if $k < 0$

(C) a point if $k = 0$

(D) a hyperbola if $k > 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, second-degree equation: $2x^2 + 3y^2 - 8x - 18y + 35 = k$. At first glance, it looks like a jumble of variables and constants.

But to a mathematician, this is not just a mess; it is a hidden geometric shape waiting to be revealed. This is the beauty of coordinate geometry—we take an algebraic expression and, through the art of manipulation, uncover the physical reality it represents.

### The Anatomy of the Equation

We start by observing the structure. We have $x^2$ and $y^2$ terms with positive coefficients, $2$ and $3$ respectively.

Because these coefficients are positive but not equal, our intuition should immediately suggest an **ellipse**. However, in the world of JEE Advanced, we must prove the nature of the locus.

The constant $k$ on the right side is the pivot point. It dictates whether this equation describes a sprawling ellipse, a singular point, or an empty set.

### The Art of Completing the Square

To see the shape clearly, we group the terms:
$$(2x^2 - 8x) + (3y^2 - 18y) + 35 = k$$

Now, we perform the classic maneuver of factoring out the leading coefficients:
$$2(x^2 - 4x) + 3(y^2 - 6y) + 35 = k$$

We transform the expressions inside the parentheses into perfect squares by adding and subtracting the necessary constants:
$$2(x^2 - 4x + 4 - 4) + 3(y^2 - 6y + 9 - 9) + 35 = k$$

Distributing the coefficients, we obtain:
$$2(x - 2)^2 - 8 + 3(y - 3)^2 - 27 + 35 = k$$

When we combine the constants $-8$, $-27$, and $35$, we find that $-8 - 27 + 35 = 0$. The constants vanish, leaving us with the elegant, simplified form:
$$2(x - 2)^2 + 3(y - 3)^2 = k$$

### The Geometric Revelation

Now, we analyze the equation $2(x - 2)^2 + 3(y - 3)^2 = k$ based on the value of $k$:

**Case 1: $k = 0$**
We have the sum of two non-negative terms equaling zero. This forces $x = 2$ and $y = 3$. The entire equation collapses into a **single point**: $(2, 3)$.

**Case 2: $k > 0$**
We can divide by $k$ to obtain the standard form:
$$\frac{(x - 2)^2}{k/2} + \frac{(y - 3)^2}{k/3} = 1$$
This is the standard equation of an **ellipse** centered at $(2, 3)$.

**Case 3: $k < 0$**
We are asking for the sum of two squares to be negative, which is impossible for real numbers. Thus, there is **no locus**.

By methodically breaking down the algebra, we have turned a daunting equation into a clear, geometric story. Remember, in JEE, the math is not just about calculation; it is about visualization.

Animated Solution for Mathematics - Conic Sections: The equation $2 x^{2} + 3 y^{2} - 8 x - 18 y + 35 = k$ represents

Select Answer:

Visualized Solution

Analyzing the General Equation

Grouping $x$ and $y$ Terms

Factoring Out Leading Coefficients

Completing the Square for $x$

Completing the Square for $y$

Simplifying the Constants

Case 1: Let $k = 0$

Case 2 & 3: When $k > 0$ and $k < 0$

Final Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Anatomy of the Equation

The Art of Completing the Square

The Geometric Revelation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation 2x2+3y2−8x−18y+35=k represents

Select Answer:

Visualized Solution

Analyzing the General Equation

Grouping .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y Terms

Factoring Out Leading Coefficients

Completing the Square for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Completing the Square for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y