Let A₁, B₁, C₁ be three points in the xy-plane. Suppose that the lines A₁C₁ and B₁C₁ are tangents to

Q: Why can't I just find the equations of the tangents and solve for intersection? It feels more 'first principles'.

You absolutely can, and that is a fantastic way to verify your work! However, in the high-pressure environment of the JEE Advanced, time is your most valuable currency. Using the $T=0$ method is the 'pro' move. It bypasses the tedious quadratic equations and gets you straight to the chord's equation in seconds. Always keep the $T=0$ tool in your belt!

Q: Wait, why is the orthocenter $(0, 0)$? Is that a coincidence?

It is not a coincidence at all! In this specific configuration, the symmetry of the parabola and the position of $C_1$ on the axis of symmetry create a beautiful geometric balance. When you see a problem involving a parabola and a point on its axis, always look for symmetry before you start calculating. It often reveals the answer before you even touch your pen.

Q: Is the chord of contact always a vertical line?

Great observation! No, it is not. The chord of contact is only vertical when the external point lies on the axis of symmetry of the parabola (the $x$-axis in this case). If you moved $C_1$ to a different point, the chord of contact would tilt. Always check the position of your external point relative to the axis of the parabola first.

Question

Let $A_1, B_1, C_1$ be three points in the $xy$-plane. Suppose that the lines $A_1C_1$ and $B_1C_1$ are tangents to the curve $y^2 = 8x$ at $A_1$ and $B_1$, respectively. If $O = (0, 0)$ and $C_1 = (-4, 0)$, then which of the following statements is(are) TRUE?

Options:

(A) The length of the line segment $OA_1$ is $4\sqrt{3}$

(B) The length of the line segment $A_1B_1$ is $16$

(C) The orthocenter of the triangle $A_1B_1C_1$ is $(0, 0)$

(D) The orthocenter of the triangle $A_1B_1C_1$ is $(1, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a problem; we are exploring the elegant architecture of conic sections. We have a parabola, $y^2 = 8x$, and an external point, $C_1(-4, 0)$.

We are drawing tangents from this point to the curve, creating a triangle $A_1B_1C_1$. It sounds simple, but within this setup lies a beautiful interplay of algebra and geometry.

### The Power of the Chord of Contact

Many students, when faced with this problem, immediately try to find the equations of the two tangents. They set up the line $y - 0 = m(x + 4)$, substitute it into the parabola, and solve for $m$. While that works, it is the long road.

We are looking for the line connecting the points of tangency, $A_1$ and $B_1$. This is the **Chord of Contact**. For any parabola $y^2 = 4ax$, the equation of the chord of contact from an external point $(x_1, y_1)$ is given by the formula $T = 0$, which is $yy_1 = 2a(x + x_1)$.

Let us apply this. Our parabola is $y^2 = 8x$, so $4a = 8$, which means $a = 2$. Our point is $C_1(-4, 0)$. Substituting these into our formula:

$$y(0) = 2(2)(x + (-4))$$

This simplifies beautifully to $0 = 4(x - 4)$, which gives us $x = 4$. Just like that, we have the equation of the line $A_1B_1$. It is a vertical line!

### Finding the Coordinates

Now that we have the line $x = 4$, finding the coordinates of $A_1$ and $B_1$ is straightforward. We intersect this line with the parabola $y^2 = 8x$:

$$y^2 = 8(4) = 32$$

$$y = \pm \sqrt{32} = \pm 4\sqrt{2}$$

So, our points are $A_1(4, 4\sqrt{2})$ and $B_1(4, -4\sqrt{2})$. With these coordinates, checking Option A is a breeze. The distance from the origin $O(0, 0)$ to $A_1(4, 4\sqrt{2})$ is:

$$OA_1 = \sqrt{(4 - 0)^2 + (4\sqrt{2} - 0)^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$$

Option A is confirmed! As for Option B, the distance $A_1B_1$ is simply the difference in $y$-coordinates: $4\sqrt{2} - (-4\sqrt{2}) = 8\sqrt{2}$. Since $8\sqrt{2} 
eq 16$, Option B is false.

### The Orthocenter Climax

Now, let us tackle the orthocenter of $	riangle A_1B_1C_1$. The orthocenter is the intersection of the altitudes.

First, consider the altitude from $C_1$ to the side $A_1B_1$. Since $A_1B_1$ is a vertical line ($x = 4$), the altitude must be a horizontal line. Since it passes through $C_1(-4, 0)$, the equation is simply $y = 0$ (the $x$-axis).

Next, we need the altitude from $B_1$ to the side $A_1C_1$. First, we find the slope of $A_1C_1$:

$$m_{A_1C_1} = \frac{4\sqrt{2} - 0}{4 - (-4)} = \frac{4\sqrt{2}}{8} = \frac{1}{\sqrt{2}}$$

The altitude must be perpendicular to this, so its slope is $m_{alt} = -\sqrt{2}$. Using the point-slope form with $B_1(4, -4\sqrt{2})$:

$$y - (-4\sqrt{2}) = -\sqrt{2}(x - 4)$$

$$y + 4\sqrt{2} = -\sqrt{2}x + 4\sqrt{2}$$

$$y = -\sqrt{2}x$$

Finally, we find the intersection of our two altitudes: $y = 0$ and $y = -\sqrt{2}x$. Substituting $y = 0$ into the second equation gives $0 = -\sqrt{2}x$, which implies $x = 0$.

The orthocenter is $(0, 0)$. Option C is true, and Option D is false. You have navigated the geometry, applied the theorems, and arrived at the truth.

Select Answer:

Visualized Solution

Parabola and Point $C_{1}$

Tangents to the Parabola

Equation of Chord of Contact

Substituting $C_{1}$ into $T = 0$

Simplifying the Equation

Finding Coordinates of $A_{1}$ and $B_{1}$

Checking Option A: Length $O A_{1}$

Calculating $O A_{1}$

Checking Option B: Length $A_{1} B_{1}$

Triangle $A_{1} B_{1} C_{1}$

Defining the Orthocenter

Altitude from $C_{1}$

Slope of Side $A_{1} C_{1}$

Equation of Altitude from $B_{1}$

Finding the Orthocenter

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of the Chord of Contact

Finding the Coordinates

The Orthocenter Climax

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Select Answer:

Visualized Solution

Parabola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Tangents to the Parabola

Equation of Chord of Contact

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​ into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Simplifying the Equation

Finding Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​

Checking Option A: Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA1​

Checking Option B: Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​B1​

Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​B1​C1​