Find the area of triangle formed by tangent to ellipse and coordinate axes

Question

Find the minimum area of the triangle formed by the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the coordinate axes.

(JEE Main 2023)

Solution — Step by Step

At the point $(a\cos\theta, b\sin\theta)$ on the ellipse, the tangent equation is:

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

x-intercept (set $y = 0$ ): $x = \frac{a}{\cos\theta}$

y-intercept (set $x = 0$ ): $y = \frac{b}{\sin\theta}$

The triangle is formed by the tangent line, the x-axis, and the y-axis. Its area is:

\text{Area} = \frac{1}{2} |x\text{-intercept}| \times |y\text{-intercept}| = \frac{1}{2} \cdot \frac{a}{|\cos\theta|} \cdot \frac{b}{|\sin\theta|}

= \frac{ab}{2|\sin\theta\cos\theta|} = \frac{ab}{|\sin 2\theta|}

The area is minimised when $|\sin 2\theta|$ is maximised, i.e., $|\sin 2\theta| = 1$ , which gives $\theta = \pi/4$ .

\text{Minimum area} = \frac{ab}{1} = \boxed{ab}

This occurs when the tangent touches the ellipse at $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ .

Why This Works

The tangent at any point on the ellipse cuts both axes, forming a right triangle with the origin. As we move the point of tangency around the ellipse, the intercepts change — when the tangent is nearly horizontal, the y-intercept is huge but the x-intercept is small, and vice versa. The product is minimised at $\theta = 45°$ where the tangent makes equal “use” of both axes.

The $\sin 2\theta$ in the denominator captures this trade-off perfectly. Since $\sin 2\theta$ oscillates between $-1$ and $1$ , the minimum area is simply $ab$ .

Alternative Method — Using AM-GM inequality

From Step 3: Area $= \frac{ab}{2\sin\theta\cos\theta}$ (in the first quadrant).

By AM-GM: $\sin\theta\cos\theta \leq \frac{\sin^2\theta + \cos^2\theta}{2} = \frac{1}{2}$

So Area $\geq \frac{ab}{2 \cdot (1/2)} = ab$ .

The answer $ab$ is elegant and worth memorising. For a circle ( $a = b = r$ ), the minimum triangle area is $r^2$ . In JEE, this result is often combined with other conditions — e.g., “if the minimum area is 6 and $a = 3$ , find $b$ .” Knowing the result $ab$ directly saves computation.

Common Mistake

Students sometimes forget to consider only the first quadrant tangent (where both intercepts are positive) and get confused with signs. For the minimum area calculation, take the absolute values of intercepts and work in the first quadrant. The symmetry of the ellipse ensures the minimum area is the same in all four quadrants.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using AM-GM inequality

Common Mistake

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