For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2)?

Q: Why do we use the 'T=0' method instead of finding the derivative and using the slope-point form?

While calculus is universal, 'T=0' is a specialized tool for conic sections. It is faster, less prone to algebraic errors, and specifically designed for second-degree curves. In the heat of a JEE exam, efficiency is just as important as accuracy.

Q: Does the 'T=0' method work for every curve?

The 'T=0' method is specifically applicable to second-degree curves (conics like circles, ellipses, parabolas, and hyperbolas). It is a powerful shortcut, but always ensure the point $(x_1, y_1)$ actually lies on the curve before applying it.

Q: What if the point $(x_1, y_1)$ was not on the curve? Would 'T=0' still work?

Great question! If the point is not on the curve, 'T=0' actually gives you the equation of the 'chord of contact' if the point is outside, or a polar line. It is a very versatile tool, but for finding a tangent, the point must lie on the curve.

Question

For which of the following curves, the line $x + \sqrt{3}y = 2\sqrt{3}$ is the tangent at the point $(\frac{3\sqrt{3}}{2}, \frac{1}{2})$?

Options:

(A) $x^2 + 9y^2 = 9$

(B) $2x^2 - 18y^2 = 9$

(C) $y^2 = \frac{1}{6\sqrt{3}}x$

(D) $x^2 + y^2 = 7$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast coordinate plane, looking at a mystery curve. You have a single point, $P\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$, and a line, $x + \sqrt{3}y = 2\sqrt{3}$, that kisses this curve exactly at $P$.

This is the essence of tangency—a moment of perfect alignment. In the world of JEE Advanced, we often encounter problems that seem to demand brute-force calculus.

### The Secret Weapon: The $T=0$ Method

Today, we are going to learn a secret weapon: the **$T=0$ method**. This is not just a formula; it is a geometric shortcut that allows us to bypass the tedious process of differentiation and slope calculation.

When you see a second-degree curve, your mind should immediately jump to the $T=0$ transformation. It is the most elegant way to find the tangent to any conic section.

The rule is simple yet profound: for any curve $S=0$, the equation of the tangent at a point $(x_1, y_1)$ is given by $T=0$. This means we replace $x^2$ with $x x_1$ and $y^2$ with $y y_1$. It is a direct substitution that feels almost like magic.

### Applying the Transformation

Let us test this on our candidate, the ellipse $x^2 + 9y^2 = 9$. We take our point $P\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ and apply the transformation.

The $x^2$ term becomes $x \left(\frac{3\sqrt{3}}{2}\right)$, and the $9y^2$ term becomes $9y \left(\frac{1}{2}\right)$. Our equation now reads:

$$x \left(\frac{3\sqrt{3}}{2}\right) + 9y \left(\frac{1}{2}\right) = 9$$

Now, let us clean this up. We multiply the entire equation by $2$ to clear the denominators:

$$3\sqrt{3}x + 9y = 18$$

### Final Verification

To match our target line $x + \sqrt{3}y = 2\sqrt{3}$, we divide the entire equation by $3\sqrt{3}$.

The first term, $\frac{3\sqrt{3}x}{3\sqrt{3}}$, simplifies beautifully to $x$. The second term, $\frac{9y}{3\sqrt{3}}$, simplifies to $\frac{3}{\sqrt{3}}y$, which is $\sqrt{3}y$.

Finally, the right side, $\frac{18}{3\sqrt{3}}$, simplifies to $\frac{6}{\sqrt{3}}$, which is $2\sqrt{3}$. The resulting equation is:

$$x + \sqrt{3}y = 2\sqrt{3}$$

It matches perfectly! This is the beauty of coordinate geometry. We did not need to calculate a single derivative; we simply used the inherent structure of the conic section to find the tangent.

Remember, the **$T=0$ method** is your best friend in the exam hall. It saves time, reduces the chance of error, and reveals the underlying symmetry of the curves you are studying.

Animated Solution for Mathematics - Conic Sections: For which of the following curves, the line $x + 3 y = 23$ is the tangent at the point $(\frac{3 3}{2}, \frac{1}{2})$ ?

Select Answer:

Visualized Solution

Visualizing the Given Information

The $T = 0$ Method

Transformation Rules for $T = 0$

Testing Option 1

Applying $T = 0$ to Option 1

Clearing the Denominators

Simplifying the Equation

Final Simplification

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Secret Weapon

Applying the Transformation

Final Verification

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For which of the following curves, the line x+3​y=23​ is the tangent at the point (233​​,21​)?