If 3x + 4y = 12√2 is a tangent to the ellipse x²/a² + y²/9 = 1, for some a ∈ R then the distance bet

Q: Wait, why do we use $c^2 = a^2m^2 + b^2$? Can't we just use the discriminant method?

You absolutely can! Setting the discriminant of the quadratic equation formed by substituting the line into the ellipse equation to zero will yield the same result. However, in the high-pressure environment of JEE, memorizing the standard tangency condition $c^2 = a^2m^2 + b^2$ saves precious minutes and reduces the chance of calculation errors.

Q: Does it matter if $a^2$ is less than $b^2$?

Great observation! If $a^2 < b^2$, the ellipse is vertically elongated (major axis along the y-axis). The tangency condition $c^2 = a^2m^2 + b^2$ remains valid for the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Just ensure you correctly identify $a^2$ and $b^2$ from the equation.

Q: Why is the distance between foci $2ae$?

The foci are located at $(\pm ae, 0)$ for a horizontal ellipse. The distance from the center $(0,0)$ to one focus is $ae$. Since there are two foci, one at $+ae$ and one at $-ae$, the total distance is $ae - (-ae) = 2ae$. It's all about the symmetry of the ellipse!

Question

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$, for some $a \in R$ then the distance between the foci of the ellipse is :

Options:

(A) $2\sqrt{5}$

(B) $2\sqrt{7}$

(C) $2\sqrt{2}$

(D) $4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to peel back the layers of a classic conic section problem. It is not just about finding a value; it is about understanding the elegant, rigid dance between a line and an ellipse.

Imagine you are standing on a coordinate plane. You have an ellipse, $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$, whose shape is partially hidden because we do not know the semi-major axis $a$.

Then, a line, $3x + 4y = 12\sqrt{2}$, comes along and kisses the ellipse at exactly one point. This is the definition of a tangent. Our mission is to find the distance between the foci of this ellipse.

### The Tangent's Transformation

Before we can interact with the ellipse, we must understand our line. The equation $3x + 4y = 12\sqrt{2}$ is in standard form, but for the purpose of tangency, we need it in the slope-intercept form, $y = mx + c$.

Moving $3x$ to the right, we get $4y = -3x + 12\sqrt{2}$. Dividing by $4$, we arrive at:

$$y = -\frac{3}{4}x + 3\sqrt{2}$$

Now, the slope $m = -\frac{3}{4}$ and the y-intercept $c = 3\sqrt{2}$ are laid bare. This is our first victory.

### The Gatekeeper Condition

Now, we invoke a powerful tool from our JEE toolkit: the condition of tangency. For any line $y = mx + c$ to be a tangent to the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, it must satisfy the identity:

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

Think of this as the gatekeeper; if the line satisfies this, it is a tangent. We know $c = 3\sqrt{2}$, $m = -\frac{3}{4}$, and from our ellipse equation, $b^2 = 9$.

Substituting these into our gatekeeper equation, we get:

$$(3\sqrt{2})^2 = a^2\left(-\frac{3}{4}ight)^2 + 9$$

### Unveiling the Ellipse

Let us perform the arithmetic with care. The square of $3\sqrt{2}$ is $9 	imes 2 = 18$. The square of $-\frac{3}{4}$ is $\frac{9}{16}$.

So, our equation becomes:

$$18 = a^2\left(\frac{9}{16}ight) + 9$$

Subtracting $9$ from both sides, we get $9 = a^2\left(\frac{9}{16}ight)$. The nines cancel out, leaving $1 = a^2\left(\frac{1}{16}ight)$, which means $a^2 = 16$. Thus, $a = 4$.

### The Foci's Secret

We are almost there. The distance between the foci of an ellipse is $2ae$. We have $a = 4$, but we need the eccentricity $e$.

The formula for eccentricity is $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting our values:

$$e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{16 - 9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$

Finally, the distance between the foci is $2ae$:

$$2ae = 2 	imes 4 	imes \frac{\sqrt{7}}{4} = 2\sqrt{7}$$

### Conclusion

Look at that result. **$2\sqrt{7}$**. It is not just a number; it is the culmination of understanding how lines and curves interact in the language of algebra.

You have navigated the transformation, applied the condition of tangency, solved for the unknown, and calculated the focal distance. This is the essence of JEE mathematics—taking a complex, abstract problem and breaking it down into a series of logical, beautiful steps.

Animated Solution for Mathematics - Conic Sections: If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

Select Answer:

Visualized Solution

Visualizing the Setup

Slope-Intercept Form

Identifying $m$ and $c$

The Condition of Tangency

Substituting Values

Solving for $a^{2}$

Finding $a$

Calculating Eccentricity $e$

Evaluating Eccentricity

Distance Between Foci

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Transformation

The Gatekeeper Condition

Unveiling the Ellipse

The Foci's Secret

Conclusion

Similar Questions

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