Tangents are drawn to the hyperbola 4x² - y² = 36 at the points P and Q. If these tangents intersect

Q: Why do we use $T=0$ for the chord of contact? Can't we just find the tangent equations and solve for their intersection?

That is a brilliant question! You absolutely could find the equations of the tangents using the slope form $y = mx \pm \sqrt{a^2m^2 - b^2}$, but that would involve solving a quadratic for $m$, which is algebraically tedious and prone to errors. The $T=0$ method is a powerful shortcut derived from the polar properties of conics. It is the 'JEE way'—elegant, fast, and less prone to calculation mistakes.

Q: Why is the height of the triangle $15$ units?

Great observation. The chord of contact $PQ$ is the horizontal line $y = -12$. The point $T$ is at $y = 3$. The vertical distance between a point at $y=3$ and a line at $y=-12$ is the absolute difference: $|3 - (-12)| = 15$. Always visualize the coordinate plane to avoid sign errors!

Q: What if the point $T$ was inside the hyperbola?

If $T$ were inside the hyperbola, you wouldn't be able to draw real tangents to it! The concept of a 'chord of contact' only exists for external points. This is a fundamental constraint of the geometry of conics.

Question

Tangents are drawn to the hyperbola $4x^2 - y^2 = 36$ at the points $P$ and $Q$. If these tangents intersect at the point $T(0, 3)$ then the area (in sq. units) of $\Delta PTQ$ is :

Options:

(A) 36\sqrt{5}

(B) 45\sqrt{5}

(C) 54\sqrt{3}

(D) 60\sqrt{3}

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at the equation $4x^2 - y^2 = 36$, do not just see numbers. See a path, a curve that defines the very essence of conic sections.

First, let us bring this equation into its standard, elegant form. By dividing the entire equation by $36$, we get:

$$\frac{x^2}{9} - \frac{y^2}{36} = 1$$

This is the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a^2 = 9$ and $b^2 = 36$.

Now, consider the point $T(0, 3)$. If we plug $x=0$ and $y=3$ into our equation, we get $4(0)^2 - (3)^2 = -9$, which is not $36$. Thus, $T$ is an external point.

### The Magic of the Chord of Contact

This is where the JEE magic happens. We could laboriously find the equations of the tangents, but there is a more beautiful path.

The line segment connecting $P$ and $Q$ is known as the **Chord of Contact**. For any conic $S=0$, the equation of the chord of contact from an external point $(x_1, y_1)$ is simply $T=0$.

In our case, the hyperbola is $4x^2 - y^2 - 36 = 0$. Using the standard substitution $x^2 ightarrow xx_1$ and $y^2 ightarrow yy_1$, the equation of the chord of contact becomes:

$$4xx_1 - yy_1 = 36$$

Substituting $x_1 = 0$ and $y_1 = 3$, we get $4x(0) - y(3) = 36$, which simplifies beautifully to $-3y = 36$, or $y = -12$. This is the horizontal line $PQ$.

### Finding the Points of Intersection

Now that we have the line $y = -12$, we need to find where it intersects the hyperbola. We substitute $y = -12$ back into the original hyperbola equation:

$$4x^2 - (-12)^2 = 36$$

This gives $4x^2 - 144 = 36$. Adding $144$ to both sides, we get $4x^2 = 180$, which means $x^2 = 45$.

Taking the square root, we find $x = \pm \sqrt{45} = \pm 3\sqrt{5}$. So, our points of contact are $P(3\sqrt{5}, -12)$ and $Q(-3\sqrt{5}, -12)$.

### The Final Area Calculation

We are now standing at the finish line. We need the area of $\Delta PTQ$. The area of a triangle is given by:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

The base is the length of the horizontal segment $PQ$. Since both points share the same $y$-coordinate, the length is simply the difference in their $x$-coordinates:

$$|3\sqrt{5} - (-3\sqrt{5})| = 6\sqrt{5}$$

The height is the vertical distance from the point $T(0, 3)$ to the line $y = -12$. This distance is $|3 - (-12)| = 15$.

Finally, the area is:

$$\frac{1}{2} 	imes 6\sqrt{5} 	imes 15 = 3\sqrt{5} 	imes 15 = 45\sqrt{5} 	ext{ sq. units.}$$

You have successfully navigated the geometry, applied the powerful $T=0$ theorem, and arrived at the solution. The final area is **$45\sqrt{5}$ sq. units**.

Animated Solution for Mathematics - Conic Sections: Tangents are drawn to the hyperbola $4 x^{2} - y^{2} = 36$ at the points $P$ and $Q$ . If these tangents intersect at the point $T (0, 3)$ then the area (in sq. units) of $Δ P T Q$ is :

Select Answer:

Visualized Solution

Identify the Hyperbola and Point $T$

Drawing the Tangents

The Chord of Contact

Applying $T = 0$

Equation of Line $P Q$

Finding Points $P$ and $Q$

Solving for $x^{2}$

Coordinates of $P$ and $Q$

Area of Triangle $P T Q$

Calculating the Base $P Q$

Calculating the Height

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Magic of the Chord of Contact

Finding the Points of Intersection

The Final Area Calculation

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Visualized Solution

Identify the Hyperbola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Drawing the Tangents

The Chord of Contact

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Finding Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Area of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PTQ

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