The axis of a parabola is along the line y = x and the distances of its vertex and focus from origin

Q: Why is the directrix $x+y=0$?

The axis is $y=x$, which has a slope of $1$. The directrix is perpendicular to the axis, so its slope must be $-1$. We know the vertex $V(1,1)$ is the midpoint of the focus $F(2,2)$ and the point of intersection $Z$ of the directrix and the axis. Using the midpoint formula, $(1,1) = \frac{Z + (2,2)}{2}$, we find $Z = (0,0)$. A line with slope $-1$ passing through $(0,0)$ is $y = -x$, or $x+y=0$.

Q: Why can't we just use the standard form $y^2=4ax$?

The standard form $y^2=4ax$ assumes the axis of the parabola is the x-axis. In this problem, the axis is tilted at a $45^\circ$ angle ($y=x$). While you could rotate the coordinate system, using the locus definition ($PF=PM$) is the most robust and elegant method for JEE Advanced problems involving rotated conics.

Q: Is there a risk of sign errors in the distance formula?

Absolutely. Always remember that the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$ is $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. When you square this, the modulus disappears, but the denominator $\sqrt{1^2+1^2} = \sqrt{2}$ becomes $2$. Don't forget to square that denominator!

Question

The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $\sqrt{2}$ and $2\sqrt{2}$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

Options:

(A) $(x+y)^2 = (x-y-2)$

(B) $(x-y)^2 = (x+y-2)$

(C) $(x-y)^2 = 4(x+y-2)$

(D) $(x-y)^2 = 8(x+y-2)$

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Parabola

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a geometric masterpiece.

When you look at a parabola, do not just see an equation. See a relationship—a perfect, unwavering balance between a point and a line. In this problem, we are given a parabola whose axis is the line $y=x$. This is our spine, our line of symmetry. Everything we do must respect this axis.

### Phase 1: Finding the Anchors

First, we must locate our landmarks. We are given that the vertex $V$ and the focus $F$ lie on the line $y=x$ in the first quadrant.

We know their distances from the origin are $\sqrt{2}$ and $2\sqrt{2}$ respectively. Since any point on the line $y=x$ can be written as $(k, k)$, the distance from the origin is simply:

$$ \sqrt{k^2 + k^2} = \sqrt{2k^2} = |k|\sqrt{2} $$

For the vertex, $\sqrt{2} = |k|\sqrt{2}$, so $k=1$. Thus, $V = (1,1)$.

For the focus, $2\sqrt{2} = |k|\sqrt{2}$, so $k=2$. Thus, $F = (2,2)$. We have our anchors; the parabola is born from these two points.

### Phase 2: The Invisible Boundary

Now, we need the directrix. This is the invisible boundary that defines the parabola's curvature.

We know the vertex $V$ is the midpoint between the focus $F$ and the intersection point $Z$ of the directrix and the axis. If $V$ is the midpoint of $F(2,2)$ and $Z(x_z, y_z)$, then:

$$ (1,1) = \frac{(2+x_z, 2+y_z)}{2} $$

Solving this gives us $Z = (0,0)$. The directrix passes through the origin and is perpendicular to the axis $y=x$.

A line perpendicular to $y=x$ (slope $1$) must have a slope of $-1$. Therefore, the equation of our directrix is $y - 0 = -1(x - 0)$, which simplifies beautifully to $x + y = 0$.

### Phase 3: The Locus of Perfection

This is the heart of the problem. A parabola is defined as the locus of a point $P(x,y)$ such that its distance to the focus $F$ is equal to its perpendicular distance to the directrix.

Mathematically, this is $PF = PM$. To make our lives easier, we square both sides: $PF^2 = PM^2$.

The distance $PF^2$ is $(x-2)^2 + (y-2)^2$. The perpendicular distance $PM$ to the line $x+y=0$ is:

$$ PM = \frac{|x+y|}{\sqrt{1^2+1^2}} = \frac{|x+y|}{\sqrt{2}} $$

Squaring this gives:

$$ PM^2 = \frac{(x+y)^2}{2} $$

### Phase 4: The Algebraic Dance

Now, we equate them:

$$ (x-2)^2 + (y-2)^2 = \frac{(x+y)^2}{2} $$

Let's expand the left side with care: $(x^2 - 4x + 4) + (y^2 - 4y + 4) = x^2 + y^2 - 4x - 4y + 8$. Now, multiply the entire equation by $2$ to clear the fraction:

$$ 2(x^2 + y^2 - 4x - 4y + 8) = (x+y)^2 $$

Expanding the right side gives $x^2 + y^2 + 2xy$. So, we have $2x^2 + 2y^2 - 8x - 8y + 16 = x^2 + y^2 + 2xy$.

Bringing all terms to one side, we get $x^2 + y^2 - 2xy - 8x - 8y + 16 = 0$. Notice the pattern? $x^2 + y^2 - 2xy$ is $(x-y)^2$.

So, $(x-y)^2 = 8x + 8y - 16$. Factoring out the $8$, we arrive at the final, elegant form:

$$ (x-y)^2 = 8(x+y-2) $$

This is not just an answer; it is the signature of the curve. You have successfully navigated the geometry, the definition, and the algebra. Take a moment to appreciate the symmetry. You are ready for the next challenge.

Animated Solution for Mathematics - Conic Sections: The axis of a parabola is along the line $y = x$ and the distances of its vertex and focus from origin are $2$ and $22$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

Select Answer:

Visualized Solution

Axis of the Parabola

Coordinates of Vertex $V$

Coordinates of Focus $F$

Directrix Intersection Point $Z$

Equation of the Directrix

Locus Definition of a Parabola

Applying the Distance Formulas

Expanding the Left Side

Expanding the Right Side

Cross Multiplication

Rearranging Terms

Final Equation of Parabola

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

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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