Let P be the point of intersection of the common tangents to the parabola y² = 12x and the hyperbola

Q: Why do we use the slope form of the tangent instead of the point form?

Great question! The point form requires you to know the point of tangency on both curves, which is unknown. The slope form $y = mx + c$ is a powerful tool because it allows us to define a family of lines and then constrain them using the condition of tangency, effectively reducing the problem to a single variable $m$.

Q: Wait, why did we reject $m^2 = -1$?

In the context of coordinate geometry for JEE Advanced, we are working in the real Cartesian plane. Since $m$ represents the slope of a line, it must be a real number. $m^2 = -1$ leads to imaginary slopes, which do not correspond to physical lines on our graph.

Q: How do I know if $P$ divides $SS'$ internally or externally?

The section formula handles this naturally! If the ratio $k$ comes out positive, it is an internal division. If it were negative, it would be external. Here, $k = 5/4$ is positive, confirming $P$ lies between $S$ and $S'$.

Q: Why is the hyperbola equation written as $x^2 - y^2/8 = 1$?

Standard form is essential. The given equation $8x^2 - y^2 = 8$ must be divided by $8$ to get the right-hand side to $1$. This reveals $A^2 = 1$ and $B^2 = 8$, which are critical for finding the eccentricity $e$ and the foci.

Q: Is there a shortcut to find the intersection of common tangents?

For a parabola and a hyperbola, the common tangents often exhibit symmetry. Notice that the parabola $y^2 = 12x$ is symmetric about the x-axis. If you find one tangent, the reflection across the x-axis is often the other. This symmetry is why the intersection point $P$ ended up on the x-axis ($y=0$)!

Question

Let P be the point of intersection of the common tangents to the parabola $y^2 = 12x$ and the hyperbola $8x^2 - y^2 = 8$. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

Options:

(A) $5:4$

(B) $14:13$

(C) $2:1$

(D) $13:11$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, two-dimensional coordinate plane. Before you, two distinct mathematical entities emerge: a parabola, $y^2 = 12x$, opening its arms wide to the right, and a hyperbola, $8x^2 - y^2 = 8$, stretching its branches toward infinity.

These curves seem independent, yet they share a secret connection—they are linked by common tangents. Today, we embark on a journey to find the point $P$, the intersection of these shared lines, and discover how it partitions the space between the hyperbola's foci.

### The Language of Tangents

To find where these curves touch, we must first speak their language. For the parabola $y^2 = 12x$, we identify the parameter $a = 3$.

The equation of any tangent to this parabola in terms of its slope $m$ is given by the elegant expression:

$$y = mx + \frac{3}{m}$$

Now, turn your gaze to the hyperbola. We must first normalize it into the standard form $\frac{x^2}{1} - \frac{y^2}{8} = 1$. Here, $A^2 = 1$ and $B^2 = 8$.

The tangent to a hyperbola is defined by $y = mx \pm \sqrt{A^2m^2 - B^2}$. Substituting our values, we get:

$$y = mx \pm \sqrt{m^2 - 8}$$

### The Condition of Harmony

For a line to be a common tangent, it must satisfy both equations simultaneously. This means the y-intercepts must be identical. We set the intercepts equal:

$$\frac{3}{m} = \pm \sqrt{m^2 - 8}$$

Squaring both sides, we obtain $\frac{9}{m^2} = m^2 - 8$. Multiplying by $m^2$ transforms this into a biquadratic equation:

$$m^4 - 8m^2 - 9 = 0$$

Factoring this, we find $(m^2 - 9)(m^2 + 1) = 0$. Since $m$ must be a real slope, we discard $m^2 = -1$ and embrace $m^2 = 9$, giving us $m = \pm 3$.

Our common tangents are revealed as $y = 3x + 1$ and $y = -3x - 1$.

### The Intersection and the Foci

Solving for the intersection $P$ of these two lines is straightforward. Setting $3x + 1 = -3x - 1$, we find $6x = -2$, or $x = -\frac{1}{3}$.

Substituting this back, we find $y = 0$. Thus, our point of interest is $P(-\frac{1}{3}, 0)$.

Now, we turn to the hyperbola's heart: its foci. With $A^2 = 1$ and $B^2 = 8$, the eccentricity is $e = \sqrt{1 + \frac{8}{1}} = 3$.

The foci are located at $(\pm Ae, 0)$, which simplifies to $S(3, 0)$ and $S'(-3, 0)$.

### The Final Partition

We are left with a simple, yet profound task: finding the ratio in which $P(-\frac{1}{3}, 0)$ divides the segment $SS'$. Using the section formula $x_P = \frac{k x_{S'} + 1 x_S}{k + 1}$, we substitute our coordinates:

$$-\frac{1}{3} = \frac{k(-3) + 1(3)}{k + 1}$$

Solving this algebraic puzzle, we find $-k - 1 = -9k + 9$, which simplifies to $8k = 10$, or $k = \frac{5}{4}$.

The point $P$ divides the segment $SS'$ in the ratio **$5:4$**.

Animated Solution for Mathematics - Conic Sections: Let P be the point of intersection of the common tangents to the parabola $y^{2} = 12 x$ and the hyperbola $8 x^{2} - y^{2} = 8$ . If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio:

Select Answer:

Visualized Solution

Analyze the Given Curves

Tangent to Parabola

Tangent to Hyperbola

Condition for Common Tangent

Solving for Slope $m$

Finding the Values of $m$

Equations of Common Tangents

Finding Intersection Point $P$

Eccentricity of Hyperbola

Coordinates of Foci $S$ and $S^{'}$

Setting up the Section Formula

Calculating the Ratio

Final Answer

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

Analyzing the Setup

The Language of Tangents

The Condition of Harmony

The Intersection and the Foci

The Final Partition

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