A common tangent T to the curves C1 : x²/4 + y²/9 = 1 and C2 : x²/42 - y²/143 = 1 does not pass thro

Q: Why do we equate $c^2$ instead of $c$ directly?

That is a fantastic question! When we write the tangent equation in slope form, we have a $\pm$ sign because the tangent could be above or below the center. By squaring both sides, we eliminate the ambiguity of the sign and focus purely on the condition that allows a line to be tangent to both curves simultaneously. It is a powerful algebraic shortcut that saves us from tracking multiple sign combinations unnecessarily.

Q: How do I know which tangent to pick if there are four possible lines?

The problem gives you a geometric constraint: the tangent must not pass through the fourth quadrant. The fourth quadrant is where $x > 0$ and $y < 0$. Once you find your four candidate lines (e.g., $y = 2x + 5, y = 2x - 5$, etc.), simply test a point in the fourth quadrant for each. If a line passes through the fourth quadrant, it is rejected. Only one line will satisfy the condition of staying out of that region entirely.

Q: Is there a faster way to find the point of contact $(x_1, y_1)$ without rearranging the equation?

Yes, there is! For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the point of contact for a tangent $y = mx + c$ is given by $x_1 = -\frac{a^2m}{c}$ and $y_1 = \frac{b^2}{c}$. Similarly, for the hyperbola, you can derive analogous formulas. However, I always recommend the method of comparing coefficients because it reinforces your understanding of the tangent equation itself, which is a fundamental skill for JEE Advanced.

Question

A common tangent $T$ to the curves $C_1 : \frac{x^2}{4} + \frac{y^2}{9} = 1$ and $C_2 : \frac{x^2}{42} - \frac{y^2}{143} = 1$ does not pass through the fourth quadrant. If $T$ touches $C_1$ at $(x_1, y_1)$ and $C_2$ at $(x_2, y_2)$, then $|2x_1 + x_2|$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a coordinate plane, looking at two distinct, elegant curves: an ellipse, $C_1$, and a hyperbola, $C_2$. They seem worlds apart, yet they share a secret connection—a common tangent line, $T$.

The ellipse is defined by:
$$\frac{x^2}{4} + \frac{y^2}{9} = 1$$
Here, $a^2 = 4$ and $b^2 = 9$.

The hyperbola is defined by:
$$\frac{x^2}{42} - \frac{y^2}{143} = 1$$
Here, $A^2 = 42$ and $B^2 = 143$. Our mission is to find a line $y = mx + c$ that is tangent to both.

### The Algebraic Bridge

To find a common tangent, we must speak the language of slopes. For any ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the condition for a line $y = mx + c$ to be a tangent is $c^2 = a^2m^2 + b^2$.

For our ellipse, this becomes:
$$c^2 = 4m^2 + 9$$

Similarly, for the hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, the condition is $c^2 = A^2m^2 - B^2$. Substituting our values, we get:
$$c^2 = 42m^2 - 143$$

Since the line $T$ is common to both, the $c^2$ values must be identical. We set them equal:
$$4m^2 + 9 = 42m^2 - 143$$

With a quick rearrangement, we find $38m^2 = 152$, which simplifies beautifully to $m^2 = 4$. Thus, the slope $m$ can be $\pm 2$.

Substituting $m^2 = 4$ back into our intercept equation, we find $c^2 = 4(4) + 9 = 25$, so $c = \pm 5$. We have four potential lines: $y = 2x + 5$, $y = 2x - 5$, $y = -2x + 5$, and $y = -2x - 5$.

### The Geometric Filter

Now, we apply the constraint: the tangent must not pass through the fourth quadrant. The fourth quadrant is the region where $x > 0$ and $y < 0$.

Let us test $y = 2x + 5$. If $x > 0$, then $y = 2x + 5$ is always positive, meaning it stays in the first and second quadrants. It avoids the fourth quadrant entirely!

This is our winner. The other lines, such as $y = 2x - 5$, will inevitably dip into the fourth quadrant for certain positive values of $x$. We have found our unique tangent: **$T: y = 2x + 5$**.

### The Points of Contact

We need the points of contact, $(x_1, y_1)$ on the ellipse and $(x_2, y_2)$ on the hyperbola. For the ellipse, the tangent at $(x_1, y_1)$ is:
$$\frac{xx_1}{4} + \frac{yy_1}{9} = 1$$

Rearranging this into slope-intercept form, we get:
$$y = -\frac{9x_1}{4y_1}x + \frac{9}{y_1}$$

Comparing this to $y = 2x + 5$, we equate the intercepts:
$$\frac{9}{y_1} = 5 \implies y_1 = \frac{9}{5}$$

Equating the slopes:
$$-\frac{9x_1}{4y_1} = 2$$
Substituting $y_1 = \frac{9}{5}$, we solve for $x_1$:
$$-\frac{9x_1}{4(9/5)} = 2 \implies -\frac{x_1}{4/5} = 2 \implies x_1 = -\frac{8}{5}$$

We repeat this logic for the hyperbola. The tangent at $(x_2, y_2)$ is:
$$\frac{xx_2}{42} - \frac{yy_2}{143} = 1$$

Rearranging gives:
$$y = \frac{143x_2}{42y_2}x - \frac{143}{y_2}$$

Comparing with $y = 2x + 5$, we get:
$$-\frac{143}{y_2} = 5 \implies y_2 = -\frac{143}{5}$$

Equating slopes:
$$\frac{143x_2}{42y_2} = 2$$
Substituting $y_2$, we find $x_2 = -\frac{84}{5}$.

### The Final Victory

We have our points: $(x_1, y_1) = (-\frac{8}{5}, \frac{9}{5})$ and $(x_2, y_2) = (-\frac{84}{5}, -\frac{143}{5})$. The question asks for $|2x_1 + x_2|$.

Substituting our values:
$$\left| 2\left(-\frac{8}{5}\right) + \left(-\frac{84}{5}\right) \right| = \left| -\frac{16}{5} - \frac{84}{5} \right| = \left| -\frac{100}{5} \right| = |-20| = 20$$

And there it is—the elegance of the result, **20**. You have successfully navigated the geometry, the algebra, and the constraints.

Enter Numerical Value:

Visualized Solution

Identify the Curves $C_{1}$ and $C_{2}$

General Tangent to Ellipse $C_{1}$

General Tangent to Hyperbola $C_{2}$

Condition for Common Tangent

Solving for Slope $m$

Calculating Intercept $c$

Selecting the Correct Tangent $T$

Point of Contact on Ellipse $C_{1}$

Finding $(x_{1}, y_{1})$

Point of Contact on Hyperbola $C_{2}$

Finding $(x_{2}, y_{2})$

Evaluating $∣2 x_{1} + x_{2} ∣$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Algebraic Bridge

The Geometric Filter

The Points of Contact

The Final Victory

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Enter Numerical Value:

Visualized Solution

Identify the Curves .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

General Tangent to Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

General Tangent to Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Condition for Common Tangent

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

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

Selecting the Correct Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x1​,y1​)

Point of Contact on Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x2​,y2​)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣2x1​+x2​∣