JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 16)
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 :
8
3
9
4
Explanation
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Equation of directrix
$$\Rightarrow y=-x$$
By definition of parabola,
$$\begin{aligned} & P M=P F \\ & \left|\frac{1+K}{\sqrt{2}}\right|=\sqrt{(1-2)^2+(K-2)^2} \\ & \frac{(1+K)^2}{2}=1+K^2+4-4 K \\ & 1+K^2+2 K=10+2 K^2-8 K \\ & K^2-10 K+9=0 \\ & (K-9)(K-1)=0 \\ & \therefore \quad K=1 \text { or } K=9 \end{aligned}$$
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