The minimum area of triangle formed by the tangent to the x²/a² + y²/b² = 1 & coordinate axes is
(a)

Q: Why do we use parametric coordinates instead of the slope-intercept form $y = mx \pm \sqrt{a^2 m^2 + b^2}$?

While the slope-intercept form is valid, it introduces a square root that can make the algebra significantly more cumbersome. Parametric coordinates $(a \cos \theta, b \sin \theta)$ allow us to express the tangent equation and the resulting intercepts in a much cleaner, trigonometric form, which is often easier to manipulate and minimize.

Q: Why does the area depend only on $a$ and $b$ and not on the position of the tangent?

The area expression $\Delta = \frac{ab}{\sin 2\theta}$ shows that the area *does* depend on $\theta$, which determines the position of the tangent. However, when we look for the *minimum* area, we find the value of $\theta$ that makes the area smallest. The result $ab$ is the minimum value, which is a constant property of the ellipse itself.

Q: Is it possible for the area to be zero?

No, the area cannot be zero. For the area to be zero, the denominator $\sin 2\theta$ would have to be infinite, which is impossible for a sine function. The tangent line will always intersect the coordinate axes at some finite distance from the origin, ensuring the triangle always exists.

Question

The minimum area of triangle formed by the tangent to the $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ & coordinate axes is

Options:

(A) $ab$ sq. units

(B) $\frac{a^2 + b^2}{2}$ sq. units

(C) $\frac{(a+b)^2}{2}$ sq. units

(D) $\frac{a^2 + ab + b^2}{3}$ sq. units

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an ellipse centered at the origin, defined by the equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

We aim to find the minimum area of the right-angled triangle formed by a tangent to this ellipse and the coordinate axes.

### The Algebraic Dance

To describe the tangent line, we utilize parametric coordinates. Let the point of contact be $(a \cos 	heta, b \sin 	heta)$.

This choice is powerful because it automatically satisfies the ellipse equation. The equation of the tangent at this point is given by:

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

### Finding the Intercepts

To find the area of the triangle, we must determine the $x$ and $y$ intercepts of the tangent line.

Setting $y = 0$ in the tangent equation yields the $x$-intercept:

$$x = \frac{a}{\cos 	heta}$$

Setting $x = 0$ in the tangent equation yields the $y$-intercept:

$$y = \frac{b}{\sin 	heta}$$

Thus, the base of the triangle is $OA = \frac{a}{\cos 	heta}$ and the height is $OB = \frac{b}{\sin 	heta}$.

### The Area Calculation

The area of the right-angled triangle is defined as $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Substituting our intercepts, we obtain:

$$\Delta = \frac{1}{2} \left( \frac{a}{\cos 	heta} ight) \left( \frac{b}{\sin 	heta} ight)$$

Using the trigonometric identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$, we simplify the expression to:

$$\Delta = \frac{ab}{\sin 2	heta}$$

### The Climax: Minimization

Since $a$ and $b$ are constants, minimizing $\Delta$ is equivalent to maximizing the denominator $\sin 2	heta$.

The maximum value of the sine function is $1$, which occurs when $2	heta = 90^\circ$ (or $	heta = 45^\circ$). Substituting this maximum value into our area expression, we find:

$$\Delta_{\min} = \frac{ab}{1} = ab$$

The **minimum area** of the triangle formed by the tangent to the ellipse and the coordinate axes is exactly **$ab$** square units.

Animated Solution for Mathematics - Conic Sections: The minimum area of triangle formed by the tangent to the $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ & coordinate axes is

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Visualized Solution

$T h e G eo m e t r i c S e t u p$

$T h e T an g e n t an d t h e T r ian g l e$

$P a r am e t r i c E q u a t i o n o f T an g e n t$

$F in d in g t h e x - in t er ce pt A$

$F in d in g t h ey - in t er ce ptB$

$A r e a o f T r ian g l e O A B$

$S u b s t i t u t in g t h e I n t er ce pt s$

$T r i g o n o m e t r i c S im pl i f i c a t i o n$

$M inimi z in g t h e A r e a$

$T h e F ina l M inim u m A r e a$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Algebraic Dance

Finding the Intercepts

The Area Calculation

The Climax

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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 minimum area of triangle formed by the tangent to the a2x2​+b2y2​=1 & coordinate axes is

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingthex−interceptA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingthey−interceptB

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

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

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

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

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