Let C be the circle of minimum area touching the parabola y=6-x 2 and the lines y=√(3) x . Then, which one of the following points lies on the circle C ?

Mathematics · Parabola

JEE Main 2024 — 6 April, Shift 1 — Question 7

Let C be the circle of minimum area touching the parabola $y=6-x^{2}$ and the lines $y=\sqrt{3}|x|$ . Then, which one of the following points lies on the circle C ?

Option A:
$(2,4)$
Correct
Option B:
$(1,2)$
Option C:
$(2,2)$
Option D:
$(1,1)$

Answer: A

Step-by-step solution

Equation of circle

$x^{2}+(y-(6-r))^{2}=r^{2}$

touches $\sqrt{3} \mathrm{x}-\mathrm{y}=0$

$\mathrm{p}=\mathrm{r}$

$\frac{|0-(6-r)|}{2}=r$

$|\mathrm{r}-6|=2 \mathrm{r}$ ;

$\mathrm{r}=2$

$\therefore$ Circle $\mathrm{x}^{2}+(\mathrm{y}-4)^{2}=4$

$(2,4)$ Satisfies this equation

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Mathematics
Chapter: Parabola
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