For which of the following curves, the line x + 3 y = 2 3 is the tangent at the point ( 3 3 2 , (1)/(2)) ?

Correct answer: C. x ^ 2 + 9 y ^ 2 = 9 m = - 1 3 , c = 2 (1)c = a ^ 2 m ^ 2 - b ^ 2 c = (9)/(2) · (1)/(3) - (1)/(2) = 1 (incorrect) (2)c = (a)/(m) = 1 24 3 - 1 3 = - (1)/(24) (incorrect) (3)c = ± a ^ 2 m ^ 2 + b ^ 2 c = ± 9 · (1)/(3) + 1 = ± 2 (correct) (4)c = ± a 1 + m ^ 2 c = ± 2 7 3 (incorrect)

JEE Main 202124 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Application of Derivatives

For which of the following curves, the line x + 3 y = 2 3 is the...

For which of the following curves, the line

x + \sqrt{3}y = 2\sqrt{3} is the tangent at the point \left(\dfrac{3\sqrt{3}}{2}, \dfrac{1}{2}\right) ?

Options

A $2{x}^{2} - 18{y}^{2} = 9$
B ${y}^{2} = \dfrac{1}{6\sqrt{3}}x$
C ${x}^{2} + 9{y}^{2} = 9$
D ${x}^{2} + {y}^{2} = 7$

Correct answer

C. ${x}^{2} + 9{y}^{2} = 9$

Step-by-step solution

$m = - \dfrac{1}{\sqrt{3}}, c = 2$

\left(1\right)c = \sqrt{{a}^{2}{m}^{2} - {b}^{2}} $$ c = \sqrt{\dfrac{9}{2} \cdot \dfrac{1}{3} - \dfrac{1}{2}} = 1 $(incorrect)\left(2\right)c = \dfrac{a}{m} = \dfrac{\dfrac{1}{24\sqrt{3}}}{\dfrac{- 1}{\sqrt{3}}} = - \dfrac{1}{24}$ (incorrect)

\left(3\right)c = \pm \sqrt{{a}^{2}{m}^{2} + {b}^{2}} $$ c = \pm \sqrt{9 \cdot \dfrac{1}{3} + 1} = \pm 2 $(correct)\left(4\right)c = \pm a\sqrt{1 + {m}^{2}}$

$c = \pm \dfrac{2\sqrt{7}}{\sqrt{3}}$ (incorrect)

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