Let for two distinct values of p the lines y = x + p touch the ellipse E : x²/4² + y²/3² = 1 at the

Q: Wait, why do we reject the negative root when calculating $p^2 = 25$? Or do we?

Great catch! We actually don't reject it. The problem states there are two distinct values of $p$. Since $p^2 = 25$, we get $p = 5$ and $p = -5$. These correspond to the two parallel tangent lines $y = x + 5$ and $y = x - 5$. Both are valid and necessary to define the quadrilateral.

Q: Why can't we just use the distance between the parallel lines to find the area?

That would only work if the quadrilateral were a rectangle or a square! Since $ABCD$ is a general parallelogram, the area is the base times the perpendicular height, or more simply, the determinant of the vectors forming the sides. Using the coordinates is the most robust method here.

Q: Is there a faster way to find the area without calculating all four coordinates?

Yes, you could use the property that the area of a parallelogram formed by tangents and secants can be related to the ellipse parameters, but in the heat of the JEE exam, calculating the coordinates is often safer and less prone to memory errors. The symmetry makes the coordinate calculation very fast anyway!

Question

Let for two distinct values of $p$ the lines $y = x + p$ touch the ellipse $E : \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$ at the points $A$ and $B$. Let the line $y = x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

Options:

(A) 36

(B) 24

(C) 48

(D) 20

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Unlocking the Ellipse

Welcome, future engineers. Today, we are not just solving a problem; we are embarking on a journey through the elegant world of conic sections.

When you look at an ellipse, do not just see an equation. See a dance of points, a balance of forces, and a perfect symmetry that, if you know how to look, will reveal the answer before you even touch your pen to paper.

### Analyzing the Setup

We begin with our ellipse, defined by the equation:

$$\frac{x^2}{16} + \frac{y^2}{9} = 1$$

Immediately, your eyes should be trained to extract the parameters. We have $a^2 = 16$ and $b^2 = 9$. This is our foundation.

The problem introduces lines $y = x + p$. Notice the slope $m = 1$. These lines are our 'gatekeepers'—they touch the ellipse at points $A$ and $B$.

### The Tangency Dance

In the JEE Advanced arena, you must avoid the brute force of substituting the line equation into the ellipse equation and solving a quadratic discriminant. Instead, rely on the condition of tangency:

$$c^2 = a^2m^2 + b^2$$

Substituting our values, we get $p^2 = 16(1)^2 + 9$, which simplifies beautifully to $p^2 = 25$. This gives us $p = \pm 5$.

We have two parallel tangents: $y = x + 5$ and $y = x - 5$. They are perfectly balanced on either side of the origin.

### Finding the Contact Points

Now, where do these lines kiss the ellipse? We need the points of contact $A$ and $B$. The standard formula for the point of contact of a line $y = mx + c$ with an ellipse is:

$$\left( -\frac{a^2m}{c}, \frac{b^2}{c} ight)$$

For $p = 5$, the point $A$ becomes:

$$\left( -\frac{16(1)}{5}, \frac{9}{5} ight) = \left( -\frac{16}{5}, \frac{9}{5} ight)$$

For $p = -5$, the point $B$ becomes:

$$\left( -\frac{16(1)}{-5}, \frac{9}{-5} ight) = \left( \frac{16}{5}, -\frac{9}{5} ight)$$

Look at these coordinates. Point $B$ is the reflection of point $A$ through the origin. This symmetry is not a coincidence; it is the inherent nature of the ellipse.

### The Intersection

Next, we have the line $y = x$ cutting through the heart of the ellipse. To find the intersection points $C$ and $D$, we substitute $y = x$ into the ellipse equation:

$$\frac{x^2}{16} + \frac{x^2}{9} = 1$$

Combining the fractions, we get:

$$\frac{9x^2 + 16x^2}{144} = 1 \Rightarrow 25x^2 = 144$$

Thus, $x^2 = \frac{144}{25}$, giving us $x = \pm \frac{12}{5}$. Since $y = x$, our points are $C\left( \frac{12}{5}, \frac{12}{5} ight)$ and $D\left( -\frac{12}{5}, -\frac{12}{5} ight)$.

### The Grand Finale

We have four points: $A, B, C,$ and $D$. Because $A$ and $B$ are symmetric about the origin, and $C$ and $D$ are symmetric about the origin, the diagonals $AB$ and $CD$ bisect each other at the origin. This confirms that $ABCD$ is a parallelogram.

In a parallelogram centered at the origin, the area is simply four times the area of the triangle formed by the origin and two adjacent vertices, such as $	riangle OAC$. The area of a triangle with vertices at $(0,0), (x_1, y_1),$ and $(x_2, y_2)$ is given by $\frac{1}{2} |x_1y_2 - x_2y_1|$.

Let's calculate the area of $	riangle OAC$ using $A\left( -\frac{16}{5}, \frac{9}{5} ight)$ and $C\left( \frac{12}{5}, \frac{12}{5} ight)$:

$$	ext{Area}(	riangle OAC) = \frac{1}{2} \left| \left( -\frac{16}{5} ight) \left( \frac{12}{5} ight) - \left( \frac{12}{5} ight) \left( \frac{9}{5} ight) ight|$$

$$	ext{Area}(	riangle OAC) = \frac{1}{2} \left| -\frac{192}{25} - \frac{108}{25} ight| = \frac{1}{2} \left| -\frac{300}{25} ight| = \frac{1}{2} 	imes 12 = 6$$

Finally, the total area of the quadrilateral $ABCD$ is $4 	imes 6 = 24$.

There it is. Through symmetry and the elegant application of coordinate geometry, we have arrived at the solution. The final area is **24**.

Select Answer:

Visualized Solution

Visualizing the Ellipse

Condition for Tangency

Solving for $p$

Point of Contact Formula

Finding Points $A$ and $B$

Intersecting Line $y = x$

Solving for $x$

Finding Points $C$ and $D$

Symmetry Observation

Area of Triangle $O A C$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Elegance

Analyzing the Setup

The Tangency Dance

Finding the Contact Points

The Intersection

The Grand Finale

Similar Questions

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is:

Let $A$ be the centre of the circle $x^{2} + y^{2} - 2 x - 4 y - 20 = 0$ . Suppose that the tangents at the points $B (1, 7)$ and $D (4, - 2)$ on the circle meet at the point $C$ . Find the area of the quadrilateral $A B C D$ .

If the tangent at a point P on the parabola $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ , then the area of the triangle PQR is: