From a point A common tangents are drawn to the circle x² + y² = a²/2 and parabola y² = 4ax. Find th

Q: Wait, why is the chord of contact a vertical line for both curves?

Great observation! The intersection point $A$ of the common tangents lies on the x-axis at $(-a, 0)$. For a circle centered at the origin, the chord of contact from a point on the x-axis is always a vertical line. Similarly, for the parabola $y^2 = 4ax$, the chord of contact from a point on the axis of the parabola (the x-axis) is also a vertical line. This symmetry is the key to simplifying the problem!

Q: Why did we reject $m^2 = -2$ when solving the quadratic?

Remember, $m$ represents the slope of a real tangent line. In the coordinate plane, the slope $m$ must be a real number. Since $m^2 = -2$ implies $m = \pm i\sqrt{2}$, which is imaginary, we must reject it. Only $m^2 = 1$ gives us the real slopes $m = 1$ and $m = -1$ that we need.

Q: Is it always a trapezium? How can I be sure?

Yes! A quadrilateral is a trapezium if it has at least one pair of parallel sides. Here, both chords of contact are vertical lines (of the form $x = k_1$ and $x = k_2$). Since all vertical lines are parallel to each other, the two chords of contact are parallel. Thus, the quadrilateral formed by these two parallel lines and the two common tangents is indeed a trapezium.

Question

From a point $A$ common tangents are drawn to the circle $x^2 + y^2 = a^2/2$ and parabola $y^2 = 4ax$. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student, to a beautiful exploration of coordinate geometry. Today, we are not just solving a problem; we are uncovering the hidden symmetry between a circle and a parabola.

Imagine you are standing in the Cartesian plane. On one side, you have a circle, $x^2 + y^2 = \frac{a^2}{2}$, a perfect, balanced shape centered at the origin. On the other, a parabola, $y^2 = 4ax$, a curve that stretches infinitely to the right.

We are tasked with finding the area of a quadrilateral formed by their common tangents and their chords of contact. It sounds daunting, but let us break it down with the precision of a master architect.

### The Algebraic Dance

To find the common tangents, we must first speak the language of the parabola. Any tangent to the parabola $y^2 = 4ax$ can be elegantly described by the slope-intercept form: 
$$y = mx + \frac{a}{m}$$

This equation is our universal key. It tells us that for any slope $m$, there exists a line that kisses the parabola at exactly one point.

But we need this line to also kiss our circle. The condition for a line $y = mx + c$ to be tangent to a circle $x^2 + y^2 = r^2$ is that the perpendicular distance from the center $(0,0)$ to the line must equal the radius $r$.

Here, $r^2 = \frac{a^2}{2}$, so $r = \frac{a}{\sqrt{2}}$. Applying the condition $c^2 = r^2(1 + m^2)$ with $c = \frac{a}{m}$, we get the beautiful equation: 
$$\left(\frac{a}{m}ight)^2 = \frac{a^2}{2}(1 + m^2)$$

Notice the magic here? The parameter $a^2$ appears on both sides. We can divide it out, leaving us with: 
$$\frac{1}{m^2} = \frac{1 + m^2}{2}$$

Cross-multiplying leads us to the quadratic in $m^2$: 
$$m^4 + m^2 - 2 = 0$$

Factoring this, we find $(m^2 + 2)(m^2 - 1) = 0$. Since $m$ must be real, we discard $m^2 = -2$ and embrace $m^2 = 1$. Thus, our slopes are $m = 1$ and $m = -1$.

### The Anchor Point

With our slopes $m = \pm 1$, the equations of our common tangents become $y = x + a$ and $y = -x - a$. Where do these two lines meet?

Solving $x + a = -x - a$ gives $2x = -2a$, or $x = -a$. Substituting this back, we find $y = 0$.

Our intersection point $A$ is $(-a, 0)$. This point is the anchor of our entire construction.

### The Vertical Boundaries

Now, we find the chords of contact from point $A$. For the circle, the chord of contact formula $T = 0$ gives: 
$$x(-a) + y(0) = \frac{a^2}{2}$$

This simplifies to $x = -\frac{a}{2}$. For the parabola, the formula $T = 0$ gives: 
$$y(0) = 2a(x - a)$$

This simplifies to $x = a$. We have two vertical lines, $x = -\frac{a}{2}$ and $x = a$. These are our parallel boundaries!

### The Trapezium Realized

We have created a trapezium. The parallel sides are the vertical segments between the tangents at $x = -\frac{a}{2}$ and $x = a$.

At $x = -\frac{a}{2}$, the $y$-values are $\pm \frac{a}{2}$, so the length is $a$. At $x = a$, the $y$-values are $\pm 2a$, so the length is $4a$.

The height is the horizontal distance between the lines: 
$$a - \left(-\frac{a}{2}ight) = \frac{3a}{2}$$

Finally, the area is: 
$$\frac{1}{2} 	imes (a + 4a) 	imes \frac{3a}{2} = \frac{15a^2}{4}$$

Take a moment to appreciate this. Through simple algebra and geometric insight, we have tamed the complexity of these curves. You have done well, student. Keep this clarity, and the next problem will be just as conquerable. The final area is **$\frac{15a^2}{4}$**.

Visualized Solution

Visualizing the Curves

Equation of Tangent to Parabola

Condition for Tangency to Circle

Setting up the Equation for $m$

Solving for the Slope $m$

Finding the Slopes

Finding the Intersection Point $A$

Chord of Contact of the Circle

Chord of Contact of the Parabola

Identifying the Quadrilateral

Finding the Vertices of the Trapezium

Calculating the Dimensions

Final Area Calculation

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

Analyzing the Setup

The Algebraic Dance

The Anchor Point

The Vertical Boundaries

The Trapezium Realized

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