Parabola
Common tangent to two curves, distance from point
nta_pyq_2023_jan
Grade 11
Question:
The distance of the point $(6, -2\sqrt{2})$ from the common tangent $y = mx + c$, $m > 0$, of the curves $x = 2y^2$ and $x = 1 + y^2$ is
Step-by-Step Solution
Key Concept: Write tangent to each curve in slope form and equate to find common tangent, then compute distance
For $y^2 = x/2$: tangent $y = mx + 1/(8m)$. For $x = 1+y^2 \Rightarrow y^2 = x-1$: substituting tangent gives discriminant = 0 condition: $m = 1/(2\sqrt{2})$. Tangent: $x - 2\sqrt{2}y + 1 = 0$. Distance from $(6,-2\sqrt{2})$: $d = |6 + 4\sqrt{2}\cdot\sqrt{2}+1|/\sqrt{1+8} = |6+8+1|/3 = 5$. Answer: (2)
Correct Answer: 5