The axis of a parabola is the line y = x and its vertex and focus are in the first quadrant at dista

Q: Wait, why is the directrix $x+y=0$? I thought the directrix was always $x = -a$ or $y = -a$.

That is a common trap! Those standard forms only apply when the axis is the x-axis or y-axis. Here, the axis is $y=x$. Since the directrix is always perpendicular to the axis, its slope must be $-1$. Because it passes through the origin (the foot of the directrix), the equation is simply $y - 0 = -1(x - 0)$, which simplifies to $x+y=0$.

Q: Why can't we just use the standard equation $Y^2 = 4AX$?

You absolutely can, but it requires a rotation of axes. You would need to define a new coordinate system where the axis of the parabola is the new X-axis. That is a valid path, but it is much more prone to calculation errors. Using the definition $PS = PM$ is the most elegant and robust method for any orientation.

Q: Why did we get two values for $k$? Does the parabola hit the line $x=1$ twice?

Yes, exactly! A parabola is an open curve. If you draw a line $x=1$ (a vertical line), it will intersect the parabola at two distinct points unless the line is tangent to the parabola. In our case, one of those points is the vertex $(1, 1)$, and the other is $(1, 9)$.

Question

The axis of a parabola is the line $y = x$ and its vertex and focus are in the first quadrant at distances $\sqrt{2}$ and $2\sqrt{2}$ units from the origin, respectively. If the point $(1, k)$ lies on the parabola, then a possible value of $k$ is :-

Options:

(A) 4

(B) 9

(C) 3

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at the line $y=x$. This is not just a line; it is the axis of our parabola, a line of symmetry that cuts the first quadrant with perfect precision.

Our journey begins by locating the two most important landmarks: the vertex $V$ and the focus $S$. We are told the vertex $V$ lies on this axis at a distance of $\sqrt{2}$ from the origin.

Since the line $y=x$ makes a $45^\circ$ angle with the positive x-axis, we can use simple trigonometry to find its coordinates:
$$V = (\sqrt{2} \cos 45^\circ, \sqrt{2} \sin 45^\circ) = (1, 1)$$

Similarly, the focus $S$ sits further along the same path at a distance of $2\sqrt{2}$, placing it at $(2, 2)$.

### The Hidden Directrix

Now, we need the directrix. A parabola is defined by its focus and its directrix. We know the vertex $V(1, 1)$ is the midpoint between the focus $S(2, 2)$ and the foot of the directrix $Z$.

A quick calculation shows that $Z$ must be at $(0, 0)$, the origin. Since the directrix is always perpendicular to the axis of the parabola, and our axis has a slope of $1$, the directrix must have a slope of $-1$.

Passing through the origin, the equation of the directrix is simply:
$$x + y = 0$$

### The Locus of Points

With the focus $S(2, 2)$ and the directrix $x+y=0$ in hand, we invoke the fundamental definition of a parabola: for any point $P(x, y)$ on the curve, the distance to the focus $PS$ must equal the perpendicular distance to the directrix $PM$.

Mathematically, $PS = PM$, or more conveniently, $PS^2 = PM^2$. The distance squared from $P(x, y)$ to $S(2, 2)$ is:
$$(x - 2)^2 + (y - 2)^2$$

The perpendicular distance from $P(x, y)$ to the line $x+y=0$ is given by the formula $\frac{|x+y|}{\sqrt{1^2+1^2}}$, so the squared distance is:
$$\frac{(x+y)^2}{2}$$

Equating these, we get the master equation:
$$(x - 2)^2 + (y - 2)^2 = \frac{(x+y)^2}{2}$$

### Solving for the Unknown

We are given that the point $(1, k)$ lies on this parabola. Substituting $x=1$ and $y=k$ into our equation, we get:
$$(1 - 2)^2 + (k - 2)^2 = \frac{(1+k)^2}{2}$$

Expanding the left side:
$$1 + (k^2 - 4k + 4) = k^2 - 4k + 5$$

Expanding the right side:
$$\frac{k^2 + 2k + 1}{2}$$

Multiplying by $2$ to clear the fraction:
$$2k^2 - 8k + 10 = k^2 + 2k + 1$$

Rearranging into a standard quadratic:
$$k^2 - 10k + 9 = 0$$

Factoring this gives $(k - 9)(k - 1) = 0$. Thus, $k$ can be $1$ or $9$.

If $k=1$, we are at the vertex. If $k=9$, we have found the other point on the parabola. The possible values for $k$ are **$1$ and $9$**.

Animated Solution for Mathematics - Conic Sections: The axis of a parabola is the line $y = x$ and its vertex and focus are in the first quadrant at distances $2$ and $22$ units from the origin, respectively. If the point $(1, k)$ lies on the parabola, then a possible value of $k$ is :-

Select Answer:

Visualized Solution

Visualizing the Parabola's Axis

Locating Vertex $V$

Locating Focus $S$

Finding the Foot of the Directrix

Equation of the Directrix

Defining the Parabola Locus

Setting up the Parabola Equation

Substituting the Point $P (1, k)$

Expanding the Left Hand Side

Expanding the Right Hand Side

Forming the Quadratic Equation

Solving the Quadratic Equation

Conclusion and Final Answer

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

Analyzing the Setup

The Hidden Directrix

The Locus of Points

Solving for the Unknown

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