For which of the following curves, the line $x+\sqrt{3 y}=2 \sqrt{3}$ is the tangent at the point $\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right)$ ?

(B) $y^{2}=\frac{1}{6 \sqrt{3}} x$

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Application of Derivatives

For which of the following curves, the line

x + 3 y = 23

is the tangent at the point

(\frac{3 3}{2}, \frac{1}{2})

Answer: (D)

The line

x + 3 y = 23

has slope

- \frac{1}{3}

and passes through

(\frac{3 3}{2}, \frac{1}{2})

(since

\frac{3 3}{2} + \frac{3}{2} = 23

). For

x^{2} + 9 y^{2} = 9

: the point satisfies

\frac{27}{4} + \frac{9}{4} = 9

✓

, and

2 x + 18 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{9 y} = - \frac{3 3 /2}{9/2} = - \frac{1}{3}

, matching the line. Options A, B, C each fail the slope test at that point, so the line is tangent only to

x^{2} + 9 y^{2} = 9

, option (D).

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option