Let the shortest distance from (a,0), a0 to the parabola y²=4x be 4. Then the equation of the circl

Q: Wait, why do we minimize $D^2$ instead of $D$?

That is a fantastic question! Minimizing the distance $D$ involves dealing with square roots, which makes the calculus messy and prone to errors. Since the distance $D$ is always non-negative, minimizing $D^2$ is mathematically equivalent to minimizing $D$. It simplifies the algebra significantly, allowing us to focus on the core physics and geometry.

Q: Why is the shortest distance always along the normal?

Think of it this way: if the line connecting the point to the curve were not the normal, you could always move slightly along the curve to get closer to the point. The only point where you cannot get any closer by moving along the curve is where the tangent is perpendicular to the line connecting the point to the curve—which is, by definition, the normal.

Q: How do we know the points $(1,0)$ and $(5,0)$ are the diameter of the circle?

The problem states the circle passes through $(5,0)$ and the focus $(1,0)$, and its center lies on the axis of the parabola (the x-axis). Since the center is on the x-axis, the line segment connecting these two points is a chord that passes through the center. Any chord passing through the center is a diameter!

Question

Let the shortest distance from $(a,0)$, $a>0$ to the parabola $y^{2}=4x$ be 4. Then the equation of the circle passing through the point $(a,0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

Options:

(A) $x^{2}+y^{2}-10x+9=0$

(B) $x^{2}+y^{2}-6x+5=0$

(C) $x^{2}+y^{2}-4x+3=0$

(D) $x^{2}+y^{2}-8x+7=0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the parabola $y^2 = 4x$. You have a point $P(a, 0)$ sitting on the x-axis.

The shortest distance from any point to a curve is always measured along the **normal** to the curve at the point of intersection. This is the fundamental geometric reality we must embrace.

### The Calculus of Minimization

To turn this into math, let us define a general point $Q$ on the parabola using parametric coordinates: $Q(t^2, 2t)$. The distance $D$ between $P(a, 0)$ and $Q(t^2, 2t)$ is given by the distance formula.

To make our lives easier, we work with the square of the distance:
$$D^2 = (t^2 - a)^2 + (2t - 0)^2$$

Expanding this, we get $D^2 = t^4 - 2at^2 + a^2 + 4t^2$. By grouping the terms, we see:
$$D^2 = t^4 + (4 - 2a)t^2 + a^2$$

This is a quadratic in terms of $t^2$. To find the minimum, we take the derivative with respect to $t^2$ and set it to zero:
$$\frac{d(D^2)}{d(t^2)} = 2(t^2) + (4 - 2a) = 0$$

Solving this gives us the critical relationship $t^2 = a - 2$.

### Solving for the Unknown

Now, we substitute $t^2 = a - 2$ back into our expression for $D^2$. The first term becomes $(a - 2 - a)^2 = (-2)^2 = 4$.

The second term becomes $4(a - 2) = 4a - 8$. Adding these together, we find:
$$D^2 = 4 + 4a - 8 = 4a - 4$$

We are given that the shortest distance $D = 4$, so $D^2 = 16$. Thus, $4a - 4 = 16$, which leads us to $4a = 20$, or **$a = 5$**. We have found our point $P(5, 0)$.

### The Circle Construction

Now, we shift gears to circle geometry. We need a circle passing through $P(5, 0)$ and the focus of the parabola $F(1, 0)$.

The center of this circle lies on the x-axis. Because the center is on the x-axis and the circle passes through two points on the x-axis, the segment connecting $(1, 0)$ and $(5, 0)$ must be the diameter.

The center $C$ is the midpoint:
$$C = \left( \frac{1+5}{2}, 0 \right) = (3, 0)$$

The radius $r$ is half the distance:
$$r = \frac{5-1}{2} = 2$$

The equation of the circle is $(x - 3)^2 + (y - 0)^2 = 2^2$. Expanding this, we get $x^2 - 6x + 9 + y^2 = 4$, which simplifies to our final answer:
**$$x^2 + y^2 - 6x + 5 = 0$$**

You have navigated the path from calculus to geometry with precision. Keep this intuition for normals close to your heart; it is a powerful weapon in your JEE arsenal.

Animated Solution for Mathematics - Conic Sections: Let the shortest distance from $(a, 0)$ , $a > 0$ to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

Select Answer:

Visualized Solution

Visualize the Parabola $y^{2} = 4 x$

The Property of Normals

Distance Formula for $D^{2}$

Minimizing $D^{2}$ using Calculus

Finding the Expression for $D^{2}$

Solving for $a$

Identify Focus $F$ and Point $P$

Circle Center $C$ and Radius $r$

Deriving the Equation of the Circle

Summary and Key Takeaway

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Calculus of Minimization

Solving for the Unknown

The Circle Construction

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Select Answer:

Visualized Solution

Visualize the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4x

The Property of Normals

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2

Minimizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2 using Calculus

Finding the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2

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

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

Circle Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r