The line 12x cos θ + 5y sin θ = 60 is tangent to which of the following curves?
(1) x² + y² = 169
(2

Q: Wait, why do we divide by 60? Is that just a random guess?

It is not a guess at all! In coordinate geometry, the 'standard form' of a line tangent to an ellipse is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$. By dividing by 60, we are normalizing the equation so that the right-hand side becomes 1, allowing us to perform a direct 'coefficient matching' with the standard form.

Q: How do I know this is an ellipse and not a hyperbola?

Great observation! The standard tangent equation for a hyperbola is $\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1$. Since our equation uses $\cos \theta$ and $\sin \theta$ with a plus sign, it perfectly matches the parametric tangent form of an ellipse.

Q: Can I solve this by finding the condition of tangency $c^2 = a^2 m^2 + b^2$?

Yes, you absolutely can! If you rewrite the line as $y = mx + c$, you will find that $m$ and $c$ are functions of $\theta$. Substituting these into the condition $c^2 = a^2 m^2 + b^2$ will lead you to the same result. However, the parametric method is much faster and more elegant for JEE exams.

Q: What if the question had a minus sign instead of a plus sign?

If the equation were $12x \cos \theta - 5y \sin \theta = 60$, the geometry would shift. You would be looking at a hyperbola, as the signs in the tangent equation for a hyperbola involve a subtraction.

Q: Is $\theta$ just an angle, or does it represent something else?

In this context, $\theta$ is the 'eccentric angle' of the point of tangency on the ellipse. It is a beautiful way to parameterize the curve, turning a complex algebraic problem into a simple trigonometric comparison.

Question

The line $12x \cos 	heta + 5y \sin 	heta = 60$ is tangent to which of the following curves?

Options:

(A) $x^2 + y^2 = 169$

(B) $144x^2 + 25y^2 = 3600$

(C) $25x^2 + 12y^2 = 3600$

(D) $x^2 + y^2 = 60$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, coordinate-geometry plane. You are presented with a line, but not just any line—a line that breathes and shifts.

The equation $12x \cos 	heta + 5y \sin 	heta = 60$ is not a static entity; it is a family of lines. As the parameter $	heta$ varies, this line sweeps across the plane, constantly kissing a specific curve.

Our mission is to identify that curve. This is the essence of JEE Advanced mathematics: seeing the hidden structure beneath the algebraic surface.

### The Art of Normalization

When you look at $12x \cos 	heta + 5y \sin 	heta = 60$, your intuition should immediately scream "standard form." In the world of conics, we love the number $1$ on the right-hand side. It acts as a universal anchor.

By dividing the entire equation by $60$, we transform our expression into:

$$\frac{12x \cos 	heta}{60} + \frac{5y \sin 	heta}{60} = \frac{60}{60}$$

Simplifying this, we arrive at the elegant form:

$$\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{12} = 1$$

Take a moment to appreciate this. We have stripped away the clutter. We now have a clear, clean equation where the variables $x$ and $y$ are weighted by trigonometric functions.

### The Mirror of Geometry

Now, we reach into our mental toolkit. Recall the standard equation of an ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

If you pick a point on this ellipse using the eccentric angle $	heta$, its coordinates are $(a \cos 	heta, b \sin 	heta)$. The equation of the tangent line at this specific point is given by the beautiful formula:

$$\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$$

Do you see the magic happening? When we place our simplified equation side-by-side with this standard tangent formula, the path forward becomes blindingly obvious. By comparing the coefficients, we see that $a = 5$ and $b = 12$.

### The Final Construction

We have found the parameters of our ellipse! The semi-major and semi-minor axes are $5$ and $12$, respectively.

Substituting these back into the standard ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get:

$$\frac{x^2}{5^2} + \frac{y^2}{12^2} = 1$$

$$\frac{x^2}{25} + \frac{y^2}{144} = 1$$

To reach the final form, we multiply through by the common denominator, $25 	imes 144 = 3600$. This yields the final equation of the curve:

$$144x^2 + 25y^2 = 3600$$

### Reflection

What we have just done is more than just algebra. We have taken a dynamic, shifting line and realized it is merely the shadow cast by a stationary ellipse.

This is the beauty of coordinate geometry—the ability to see the static truth behind the moving parts. Whenever you see $\sin 	heta$ and $\cos 	heta$ in a linear equation, remember: you are likely looking at a tangent, and the ellipse is waiting for you to uncover it.

Animated Solution for Mathematics - Conic Sections: The line $12 x cos θ + 5 y sin θ = 60$ is tangent to which of the following curves?

Select Answer:

Visualized Solution

Given Line Equation

Standardizing the Equation

Dividing by 60

Simplifying Fractions

Standard Tangent to Ellipse

Comparing Coefficients

Constructing the Ellipse

Final General Form

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Art of Normalization

The Mirror of Geometry

The Final Construction

Reflection

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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; } The line 12xcosθ+5ysinθ=60 is tangent to which of the following curves?