The line passing through the extremity A of the major axis and extremity B of the minor axis of the

Q: Wait, why is the auxiliary circle $x^2 + y^2 = 9$ and not something else?

Great question! The auxiliary circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is defined as the circle with the major axis as its diameter. Since our ellipse is $\frac{x^2}{3^2} + \frac{y^2}{1^2} = 1$, the semi-major axis $a$ is $3$. Thus, the circle is centered at the origin with radius $a=3$, giving us $x^2 + y^2 = 3^2 = 9$.

Q: How do I know which intersection point is $M$?

When you solve the quadratic equation $5x^2 - 3x - 36 = 0$, you get two roots: $x=3$ and $x=-\frac{12}{5}$. The root $x=3$ corresponds to point $A(3,0)$, which is where the line $AB$ starts. Therefore, the other root, $x=-\frac{12}{5}$, must be the $x$-coordinate of the point $M$ where the line intersects the circle.

Q: Can I use the determinant formula for area instead of base-height?

Absolutely! The determinant formula $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ works for any triangle. Since one vertex is the origin $(0,0)$, it simplifies to $\frac{1}{2} |x_A y_M - x_M y_A|$. Both methods will lead you to the same elegant result of $\frac{27}{10}$.

Question

The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2 + 9y^2 = 9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A, M$ and the origin $O$ is

Options:

(A) $31/10$

(B) $29/10$

(C) $21/10$

(D) $27/10$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE mastery! Today, we are going to peel back the layers of a beautiful coordinate geometry problem.

We start with the equation $x^2 + 9y^2 = 9$. At first glance, it looks like a simple curve, but it is actually an ellipse waiting to be understood.

To see its true nature, we must bring it into the standard form:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

By dividing the entire equation by $9$, we get:

$$ \frac{x^2}{9} + \frac{y^2}{1} = 1 $$

This reveals that our semi-major axis $a$ is $3$ and our semi-minor axis $b$ is $1$. Imagine this ellipse stretched along the $x$-axis, reaching from $-3$ to $3$, and hugging the $y$-axis from $-1$ to $1$.

### The Line and the Auxiliary Circle

We are interested in the line passing through the extremity of the major axis $A(3,0)$ and the extremity of the minor axis $B(0,1)$. Using the intercept form of a line, $\frac{x}{a} + \frac{y}{b} = 1$, we immediately find the equation of our line $AB$:

$$ \frac{x}{3} + \frac{y}{1} = 1 $$

Rearranging this, we get $y = 1 - \frac{x}{3}$.

Now, consider the auxiliary circle. This is the circle that encompasses the ellipse, defined by the major axis as its diameter. Since $a=3$, the equation of this circle is $x^2 + y^2 = 9$.

### The Intersection: Where Paths Cross

To find the point $M$ where the line $AB$ meets the auxiliary circle, we must solve the system of equations formed by the line and the circle. We substitute $y = 1 - \frac{x}{3}$ into $x^2 + y^2 = 9$:

$$ x^2 + \left( 1 - \frac{x}{3} ight)^2 = 9 $$

Expanding this carefully, we get:

$$ x^2 + \left( 1 - \frac{2x}{3} + \frac{x^2}{9} ight) = 9 $$

Combining the terms, we have:

$$ \frac{10x^2}{9} - \frac{2x}{3} - 8 = 0 $$

Multiplying by $9$ to clear the fractions, we arrive at $10x^2 - 6x - 72 = 0$, which simplifies to:

$$ 5x^2 - 3x - 36 = 0 $$

### The Algebraic Triumph

Now, we factor this quadratic equation. We are looking for two numbers that multiply to $5 	imes (-36) = -180$ and add to $-3$. These numbers are $-15$ and $12$.

So, we write:

$$ 5x^2 - 15x + 12x - 36 = 0 $$

This factors beautifully into $5x(x - 3) + 12(x - 3) = 0$, or $(x - 3)(5x + 12) = 0$. The roots are $x = 3$ and $x = -\frac{12}{5}$.

We know $x=3$ corresponds to point $A$. Thus, the $x$-coordinate of $M$ is $-\frac{12}{5}$. Plugging this back into our line equation:

$$ y = 1 - \frac{-\frac{12}{5}}{3} = 1 + \frac{4}{5} = \frac{9}{5} $$

So, the coordinates of $M$ are $\left( -\frac{12}{5}, \frac{9}{5} ight)$.

### The Final Area

We have reached the final stage. We need the area of $	riangle AOM$ with vertices $O(0,0)$, $A(3,0)$, and $M\left( -\frac{12}{5}, \frac{9}{5} ight)$.

Using the base-height method, the base $OA$ lies on the $x$-axis with length $3$. The height is the perpendicular distance from $M$ to the $x$-axis, which is simply the $y$-coordinate of $M$, $\frac{9}{5}$.

The area is:

$$ 	ext{Area} = \frac{1}{2} 	imes 3 	imes \frac{9}{5} = \frac{27}{10} $$

And there it is! A perfect, elegant result. You have successfully navigated the geometry, the algebra, and the final calculation. The final area is **$\frac{27}{10}$** square units.

Select Answer:

Visualized Solution

The Given Ellipse

Standard Form of the Ellipse

Identifying Extremities $A$ and $B$

Equation of Line $A B$

The Auxiliary Circle

Finding Intersection Point $M$

Expanding the Equation

Solving the Quadratic Equation

Coordinates of Point $M$

Setting up Area of $△ A O M$

Calculating the Final Area

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

Analyzing the Setup

The Line and the Auxiliary Circle

The Intersection

The Algebraic Triumph

The Final Area

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