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 :

• A. 8
• B. 4
• C. 3
• D. 9

Answer 1

Answer: (D)

9

Answer 2

The axis of the parabola is $y=x$. The vertex $V$ and focus $F$ lie on this line and are in the first quadrant. The distance of the vertex from the origin $(0,0)$ is $\sqrt{2}$. If $V=(a,a)$ with $a>0$, then $\sqrt{a^2+a^2} = \sqrt{2} \implies a\sqrt{2} = \sqrt{2} \implies a=1$. So, the vertex is $V=(1,1)$. The distance of the focus from the origin is $2\sqrt{2}$. If $F=(b,b)$ with $b>0$, then $\sqrt{b^2+b^2} = 2\sqrt{2} \implies b\sqrt{2} = 2\sqrt{2} \implies b=2$. So, the focus is $F=(2,2)$.

The vertex $V(1,1)$ is the midpoint between the focus $F(2,2)$ and the point $D$ where the directrix intersects the axis. Let $D=(d,d)$. $V = \left(\frac{F_x+D_x}{2}, \frac{F_y+D_y}{2}\right) \implies (1,1) = \left(\frac{2+d}{2}, \frac{2+d}{2}\right) \implies 2=2+d \implies d=0$. So, $D=(0,0)$. The directrix is perpendicular to the axis $y=x$ (which has a slope of 1) and passes through $(0,0)$. The slope of the directrix is $-1$. The equation of the directrix is $y-0 = -1(x-0) \implies y=-x \implies x+y=0$.

By the definition of a parabola, any point $P(x,y)$ on the parabola is equidistant from the focus and the directrix. Let $P=(1,k)$ be a point on the parabola. The distance from $P$ to the focus $F(2,2)$ is $PF = \sqrt{(1-2)^2 + (k-2)^2} = \sqrt{(-1)^2 + (k-2)^2} = \sqrt{1 + (k-2)^2}$. The distance from $P$ to the directrix $x+y=0$ is $PD = \frac{|1+k|}{\sqrt{1^2+1^2}} = \frac{|1+k|}{\sqrt{2}}$.

Equating the distances: $\sqrt{1 + (k-2)^2} = \frac{|1+k|}{\sqrt{2}}$ Squaring both sides: $1 + (k-2)^2 = \frac{(1+k)^2}{2}$ $1 + (k^2 - 4k + 4) = \frac{1 + 2k + k^2}{2}$ $k^2 - 4k + 5 = \frac{k^2 + 2k + 1}{2}$ Multiply by 2: $2(k^2 - 4k + 5) = k^2 + 2k + 1$ $2k^2 - 8k + 10 = k^2 + 2k + 1$ Rearranging into a quadratic equation: $k^2 - 10k + 9 = 0$ Factoring the quadratic: $(k-1)(k-9) = 0$ This gives two possible values for $k$: $k=1$ or $k=9$. The point $(1,1)$ is the vertex, which must lie on the parabola. The point $(1,9)$ is also a valid point on the parabola. Looking at the options provided: (a) 8, (b) 4, (c) 3, (d) 9. The value $k=9$ is present in the options.