If the length of the latus rectum of a parabola, whose focus is (a, a) and the tangent at its vertex

Q: Wait, why is the distance from the focus to the tangent at the vertex equal to $a$ in the standard form $y^2 = 4ax$? I am confused about the general case.

Great observation! In the standard parabola $y^2 = 4ax$, the vertex is at $(0,0)$ and the tangent at the vertex is the y-axis ($x=0$). The focus is at $(a,0)$. The distance from $(a,0)$ to $x=0$ is indeed $a$. Since the latus rectum is $4a$, the distance is exactly $1/4$ of the latus rectum. This property is invariant under rotation and translation, which is why it holds true for any parabola!

Q: Why can't I just find the equation of the parabola first? It seems like a lot of work to use the distance property.

You absolutely could! You would need to find the axis of the parabola (which is perpendicular to the tangent at the vertex and passes through the focus), find the vertex, and then use the definition of the parabola. However, the beauty of competitive exams like JEE is recognizing these 'geometric shortcuts.' Using the distance property saves you from tedious algebraic expansion and potential sign errors.

Q: Does the sign of $a$ matter here? The question asks for $|a|$.

The absolute value bars in the distance formula $|Ax_1 + By_1 + C|$ naturally handle the sign for us. Since the distance must be positive, $|a|$ is the correct magnitude. Whether $a$ is positive or negative, the geometric distance remains the same.

Q: Is the tangent at the vertex always perpendicular to the axis of the parabola?

Yes, always! The axis of the parabola is the line of symmetry. The tangent at the vertex is perpendicular to this axis. This is a defining characteristic of the vertex of a parabola.

Q: I keep getting $\sqrt{2}$ in the denominator. Is that correct?

Yes, it is! When you calculate the distance from $(a, a)$ to the line $x + y - a = 0$, the denominator is $\sqrt{1^2 + 1^2} = \sqrt{2}$. You are on the right track—don't let the radical intimidate you!

Question

If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x + y = a$, is 16, then $|a|$ is equal to :

Options:

(A) $2\sqrt{2}$

(B) $2\sqrt{3}$

(C) $4\sqrt{2}$

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty coordinate plane. You are given a single point, the focus $S(a, a)$, and a line, the tangent at the vertex $x + y = a$.

At first glance, this looks like a puzzle of coordinates and lines, but beneath the surface lies a beautiful, rigid geometric structure. In the world of JEE Advanced, we don't just solve for variables; we uncover the hidden symmetries of the conic sections.

### The Hidden Symmetry

Many students rush to write the general equation of a parabola, involving a second-degree polynomial in $x$ and $y$. While that is a valid path, it is often a path through a thick forest of algebra.

Instead, let us look for the 'soul' of the parabola. We know that for any parabola, the distance $d$ from the focus to the tangent at the vertex is a fundamental constant related to the latus rectum ($LR$).

Specifically, the latus rectum is exactly four times this distance:

$$LR = 4 	imes d$$

This is not just a formula; it is a geometric truth that connects the focus to the very 'tip' of the parabola.

### The Power of the Distance Formula

We are given that the length of the latus rectum is $16$. Using our elegant property, we can immediately find the distance $d$:

$$d = \frac{LR}{4} = \frac{16}{4} = 4$$

Now, we turn to the coordinate geometry. We have the focus $S(a, a)$ and the line $x + y - a = 0$.

The perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by the classic formula:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

I know that seeing variables like $a$ inside the formula can feel daunting, but take a deep breath. Let us substitute our values carefully. With $A=1$, $B=1$, and $C=-a$, the distance becomes:

$$d = \frac{|1(a) + 1(a) - a|}{\sqrt{1^2 + 1^2}}$$

### The Final Unveiling

Look at the numerator: $|a + a - a| = |a|$. It simplifies so beautifully!

The denominator is simply $\sqrt{1 + 1} = \sqrt{2}$. We are left with a simple, clean expression:

$$d = \frac{|a|}{\sqrt{2}}$$

We have already established that $d = 4$. By equating these two, we get:

$$\frac{|a|}{\sqrt{2}} = 4$$

Multiplying both sides by $\sqrt{2}$, we arrive at our destination:

**$|a| = 4\sqrt{2}$**

### Reflection

Think about what we just did. We didn't need to find the equation of the axis, nor did we need to find the coordinates of the vertex.

By relying on the geometric properties of the parabola, we bypassed the complexity and arrived at the truth through logic. This is the essence of JEE Advanced preparation—finding the most elegant path through the maze.

Keep this property in your toolkit; it is a powerful weapon for any conic section problem you encounter.

Animated Solution for Mathematics - Conic Sections: If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x + y = a$ , is 16, then $∣ a ∣$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

The Parabola Property

Calculating the Distance $d$

The Distance Formula

Substituting the Values

Simplifying the Numerator

Simplifying the Denominator

Equating to the Known Distance

Solving for $∣ a ∣$

Summary and Takeaway

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Hidden Symmetry

The Power of the Distance Formula

The Final Unveiling

Reflection

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