1. Introduction: Unveiling a Fundamental Property of Ellipses

This problem delves into a classic geometric property of ellipses that is frequently tested in JEE. We are asked to identify a point that lies on the locus of the foot of the perpendicular drawn from any of the ellipse's foci to any of its tangents.

Key Concept: The Auxiliary Circle as the Pedal Locus

The most crucial piece of information for this problem is a well-known geometric property: The locus of the foot of the perpendicular drawn from either focus of an ellipse to any tangent is its auxiliary circle.

For a standard ellipse given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

• The semi-major axis is $a$.
• The semi-minor axis is $b$.
• The equation of its auxiliary circle is $x^2 + y^2 = a^2$. This is a circle centered at the origin with a radius equal to the semi-major axis length, $a$.

Why is this property important? Understanding and memorizing this property can save significant time in competitive exams like JEE, as it allows for a direct solution without needing to perform a full derivation during the test. However, we will also provide a detailed derivation later for a deeper understanding.

2. Applying the Key Concept to the Given Ellipse

First, let's analyze the given equation of the ellipse: ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$

To apply the key concept, we need to identify the values of $a^2$ and $b^2$. We compare this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

From the comparison, we find:

• $a^2 = 4$
• $b^2 = 2$

Since $a^2 > b^2$ (i.e., $4 > 2$), the major axis of this ellipse lies along the x-axis. From $a^2=4$, we get the semi-major axis length $a = \sqrt{4} = 2$.

Now, using the key concept, the locus of the foot of the perpendicular from any focus to any tangent is the auxiliary circle. The equation of the auxiliary circle is $x^2 + y^2 = a^2$.

Substituting the value of $a^2 = 4$ into this equation: $x^2 + y^2 = 4$ This is the equation of the required locus. It represents a circle centered at the origin with a radius of $2$.

3. Verifying the Options

Our task now is to find which of the given points lies on this locus, i.e., which point $(x, y)$ satisfies the equation $x^2 + y^2 = 4$. We will substitute the coordinates of each option into this equation.

• (A) $(-1, \sqrt{3})$: Substitute $x = -1$ and $y = \sqrt{3}$ into $x^2 + y^2$: $(-1)^2 + (\sqrt{3})^2 = 1 + 3 = 4$ Since $4 = 4$, this point lies on the locus.

• (B) $(-2, \sqrt{3})$: Substitute $x = -2$ and $y = \sqrt{3}$ into $x^2 + y^2$: $(-2)^2 + (\sqrt{3})^2 = 4 + 3 = 7$ Since $7 \neq 4$, this point does not lie on the locus.

• (C) $(-1, \sqrt{2})$: Substitute $x = -1$ and $y = \sqrt{2}$ into $x^2 + y^2$: $(-1)^2 + (\sqrt{2})^2 = 1 + 2 = 3$ Since $3 \neq 4$, this point does not lie on the locus.

• (D) $(1, 2)$: Substitute $x = 1$ and $y = 2$ into $x^2 + y^2$: $(1)^2 + (2)^2 = 1 + 4 = 5$ Since $5 \neq 4$, this point does not lie on the locus.

From the verification, only option (A) satisfies the equation of the locus.

Therefore, the correct answer is (A).

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4. Detailed Derivation of the Locus (for a Deeper Understanding)

While memorizing the property is efficient, understanding why it holds true is fundamental for a strong grasp of coordinate geometry. Let's derive this property step-by-step.

Our Goal: To find the locus of a point $P(h, k)$ which is the foot of the perpendicular from a focus $S$ to any tangent of the ellipse.

Step 4.1: Identify the Foci of the Ellipse The given ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$, so $a^2=4$ and $b^2=2$. To find the foci, we first need to calculate the eccentricity $e$. For an ellipse with major axis along the x-axis, $b^2 = a^2(1 - e^2)$. Rearranging for $e^2$: $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}$ So, the eccentricity is $e = \frac{1}{\sqrt{2}}$. The coordinates of the foci are $(\pm ae, 0)$. $ae = 2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ Thus, the foci are $S_1(\sqrt{2}, 0)$ and $S_2(-\sqrt{2}, 0)$. We can pick either focus for our derivation; let's choose $S_1(\sqrt{2}, 0)$.

Step 4.2: Write the Equation of a General Tangent to the Ellipse The general equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is given by: $y = mx \pm \sqrt{a^2m^2 + b^2}$ Substituting $a^2=4$ and $b^2=2$ for our ellipse: $y = mx \pm \sqrt{4m^2 + 2}$ Let's choose the positive sign for the constant term without loss of generality, as the locus will be the same due to symmetry. Let this tangent line be $L_T$: $L_T: y - mx = \sqrt{4m^2 + 2} \quad \cdots (1)$ This equation represents any tangent to the ellipse, depending on the value of $m$.

Step 4.3: Write the Equation of the Perpendicular Line from the Focus to the Tangent We need to find the foot of the perpendicular from the focus $S_1(\sqrt{2}, 0)$ to the tangent $L_T$. If the tangent $L_T$ has a slope $m$, then any line perpendicular to it will have a slope $m_{\perp} = -\frac{1}{m}$ (assuming $m \neq 0$). This perpendicular line ($L_{\perp}$) passes through the focus $S_1(\sqrt{2}, 0)$. Using the point-slope form $y - y_1 = m_{\perp}(x - x_1)$: $y - 0 = -\frac{1}{m}(x - \sqrt{2})$ To eliminate the fraction, multiply by $m$: $my = -(x - \sqrt{2})$ $my = -x + \sqrt{2}$ Rearranging, we get the equation of the perpendicular line $L_{\perp}$: $L_{\perp}: x + my = \sqrt{2} \quad \cdots (2)$

Step 4.4: Find the Locus of the Foot of the Perpendicular $(h, k)$ Let the foot of the perpendicular be $P(h, k)$. Since $P(h, k)$ lies on both the tangent line $L_T$ and the perpendicular line $L_{\perp}$, its coordinates must satisfy both equations:

1. From equation (1), replacing $(x, y)$ with $(h, k)$: $k - mh = \sqrt{4m^2 + 2} \quad \cdots (A)$

2. From equation (2), replacing $(x, y)$ with $(h, k)$: $h + mk = \sqrt{2} \quad \cdots (B)$

Our objective is to find the relationship between $h$ and $k$ by eliminating the parameter $m$. A common and effective strategy when you have expressions of the form $(k-mh)$ and $(h+mk)$ is to square both equations and then add them. This often leads to the cancellation of cross terms and simplification involving $(h^2+k^2)$ and $(1+m^2)$.

Squaring equation (A): $(k - mh)^2 = (\sqrt{4m^2 + 2})^2$ $k^2 - 2mhk + m^2h^2 = 4m^2 + 2 \quad \cdots (A')$

Squaring equation (B): $(h + mk)^2 = (\sqrt{2})^2$ $h^2 + 2mhk + m^2k^2 = 2 \quad \cdots (B')$

Now, add equation (A') and equation (B'). Notice how the terms involving $2mhk$ will cancel out: $(k^2 - 2mhk + m^2h^2) + (h^2 + 2mhk + m^2k^2) = (4m^2 + 2) + 2$ $k^2 + m^2h^2 + h^2 + m^2k^2 = 4m^2 + 4$ Group terms with $h^2$ and $k^2$ on the left side: $(h^2 + k^2) + m^2(h^2 + k^2) = 4(m^2 + 1)$ Factor out $(h^2 + k^2)$ on the left side: $(h^2 + k^2)(1 + m^2) = 4(m^2 + 1)$ Since $m^2+1$ cannot be zero, we can divide both sides by $(m^2 + 1)$: $h^2 + k^2 = 4$ Replacing $(h, k)$ with $(x, y)$ to represent the general point on the locus, we get: $x^2 + y^2 = 4$ This confirms that the locus is indeed a circle centered at the origin with a radius of $2$, which is the auxiliary circle of the given ellipse ($a^2=4 \implies a=2$).

Common Mistake to Avoid:

• Algebraic Errors: Be very careful when squaring binomials, especially with signs, and when collecting like terms.
• Dividing by Zero: Ensure that when you divide by an expression like $(1+m^2)$, it is not zero. In this case, $m^2+1$ is always positive, so it's safe to divide.
• Special Cases (m=0 or m undefined): The derivation assumed $m \neq 0$. If $m=0$, the tangent is $y=\pm b$. The perpendicular from $(\pm ae, 0)$ is $x=\pm ae$. The intersection is $(\pm ae, \pm b)$, which is not on the auxiliary circle. This is because the slope-intercept form $y=mx \pm \sqrt{a^2m^2+b^2}$ does not cover vertical tangents ($x=\pm a$). However, the property holds for all tangents. If $x=\pm a$ are vertical tangents, the foot of the perpendicular from focus $(\pm ae, 0)$ to $x=\pm a$ is $(\pm a, 0)$, which lies on $x^2+y^2=a^2$. The derivation using $y=mx \pm \sqrt{a^2m^2+b^2}$ implicitly handles the general case, and the final equation $x^2+y^2=a^2$ covers all points.

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5. Summary and Key Takeaway

This problem beautifully illustrates a fundamental geometric property of ellipses: the locus of the foot of the perpendicular from a focus to any tangent of an ellipse is its auxiliary circle. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, this auxiliary circle is $x^2 + y^2 = a^2$.

For JEE preparation, it is highly recommended to:

1. Memorize this property for quick problem-solving.
2. Understand its derivation to build a strong conceptual foundation and handle variations of the problem.

By applying this property