If the line y = mx + c is a common tangent to the hyperbola x²/100 - y²/64 = 1 and the circle x² + y

Q: Wait, why is the hyperbola condition $c^2 = a^2m^2 - b^2$ and not $c^2 = a^2m^2 + b^2$?

That is a classic trap! It is easy to confuse the ellipse and hyperbola formulas. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the condition is $c^2 = a^2m^2 + b^2$. However, for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the 'minus' sign in the standard equation carries over to the tangency condition. Always remember: the hyperbola's 'subtraction' nature dictates its tangency condition.

Q: Can I just substitute $y = mx + c$ into the curve equations and set the discriminant to zero?

Absolutely! That is the 'brute force' method, and it works every time. However, in JEE Advanced, time is your most valuable currency. Memorizing the tangency conditions is like having a shortcut key on your keyboard. It saves you from messy quadratic expansions and potential calculation errors. Use the conditions whenever possible!

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

Great question! The tangency conditions are derived in terms of $c^2$ because the line $y = mx + c$ can be a tangent with either a positive or negative intercept (i.e., $\pm c$). By working with $c^2$, we capture both possibilities at once, simplifying the algebra significantly. It is a more robust approach.

Question

If the line $y = mx + c$ is a common tangent to the hyperbola $\frac{x^2}{100} - \frac{y^2}{64} = 1$ and the circle $x^2 + y^2 = 36$, then which one of the following is true?

Options:

(A) $5m = 4$

(B) $8m + 5 = 0$

(C) $c^2 = 369$

(D) $4c^2 = 369$

Sigma Percentile · Accepted Answer

### The Geometry of Connection

Imagine you are standing on the coordinate plane, looking at two distinct, elegant shapes. On one hand, you have the hyperbola

$$\frac{x^2}{100} - \frac{y^2}{64} = 1$$

a wild, untamed curve that stretches toward infinity. On the other, you have the circle $x^2 + y^2 = 36$, a perfect, symmetric guardian centered at the origin.

Our mission is to find a single, straight line, $y = mx + c$, that acts as a bridge between them—a common tangent that kisses both curves perfectly. This is not just an algebraic exercise; it is a search for harmony between two different geometric worlds.

### The Hyperbola's Secret

Every curve has a secret language of tangency. For our hyperbola, the condition for a line $y = mx + c$ to be a tangent is governed by the elegant formula:

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

By comparing our given hyperbola

$$\frac{x^2}{100} - \frac{y^2}{64} = 1$$

with the standard form

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

we immediately identify that $a^2 = 100$ and $b^2 = 64$.

Substituting these values, we unlock the first constraint: $c^2 = 100m^2 - 64$. Let us hold onto this as our first pillar of truth.

### The Circle's Constraint

Now, we turn our attention to the circle $x^2 + y^2 = 36$. The condition for tangency here is even more intuitive: the perpendicular distance from the center to the line must equal the radius.

This leads us to the condition:

$$c^2 = r^2(1 + m^2)$$

With our circle, $r^2 = 36$, so our second constraint becomes $c^2 = 36(1 + m^2)$. We now have two separate conditions for the same line.

The beauty of this problem lies in the fact that for the line to be a common tangent, it must satisfy both conditions simultaneously.

### The Grand Unification

This is the moment of synthesis. Since the line is a common tangent, the $c^2$ from the hyperbola must be identical to the $c^2$ from the circle.

We equate them:

$$100m^2 - 64 = 36(1 + m^2)$$

We have successfully reduced a complex geometric problem into a single-variable algebraic equation. Now, we expand the right side: $100m^2 - 64 = 36 + 36m^2$.

By grouping the $m^2$ terms on the left and the constants on the right, we get $100m^2 - 36m^2 = 36 + 64$, which simplifies to $64m^2 = 100$. Thus,

$$m^2 = \frac{100}{64} = \frac{25}{16}$$

### The Algebraic Finale

We are almost there. We have the slope squared, and now we need the intercept squared. Substituting $m^2 = \frac{25}{16}$ back into our circle condition, $c^2 = 36(1 + m^2)$, we get:

$$c^2 = 36\left(1 + \frac{25}{16}ight)$$

Simplifying the bracket, we have:

$$c^2 = 36\left(\frac{16 + 25}{16}ight) = 36\left(\frac{41}{16}ight)$$

Reducing the fraction by dividing $36$ and $16$ by $4$, we obtain:

$$c^2 = \frac{9 	imes 41}{4} = \frac{369}{4}$$

Finally, cross-multiplying gives us **$4c^2 = 369$**. We have arrived at the truth. This result is not just a number; it is the mathematical signature of the common tangent that connects these two beautiful curves.

Animated Solution for Mathematics - Conic Sections: If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

Select Answer:

Visualized Solution

Visualizing the Setup

Condition for Hyperbola

Applying to our Hyperbola

Condition for Circle

Applying to our Circle

Equating the Conditions

Expanding the Equation

Grouping Terms

Solving for $m^{2}$

Finding $c^{2}$

Simplifying the Bracket

Reducing the Fraction

Further Simplification

Final Calculation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Connection

The Hyperbola's Secret

The Circle's Constraint

The Grand Unification

The Algebraic Finale

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