Tangent & Normal – JEE Main Previous Year Question with Solutions
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Q. The equation of the tangent to the curve $y=x+\frac{4}{x^{2}},$ that is parallel to the x-axis, is :-(1) y = 0                (2) y = 1                 (3) y = 2                    (4) y = 3 [AIEEE-2010]

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Sol. (4)

Q. The intercepts on x-axis made by tangents to the curve, $\mathrm{y}=\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}, \mathrm{x} \in \mathrm{R}$ which are parallel to the line y = 2x, are equal to$(1) \pm 1$ $(2) \pm 2$ (3) $\pm 3$ (4) $\pm 4$ [JEE-MAIN 2013]

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Sol. (1)

Q. The normal to the curve, $x^{2}+2 x y-3 y^{2}=0,$ at $(1,1):$(1)meets the curve again in the third quadrant(2) meets the curve again in the fourth quadrant(3) does not meet the curve again(4) meets the curve again in the second quadrant [JEE-MAIN 2015]

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Sol. (2)

Q. Consider $f(x)=\tan ^{-1}(\sqrt{\frac{1+\sin x}{1-\sin x}}), x \in\left(0, \frac{\pi}{2}\right) .$ A normal to $y=f(x)$ at $x=\frac{\pi}{6}$ also passes through the point :( 1)$\left(\frac{\pi}{4}, 0\right)$b(2) (0, 0)(3) $\left(0, \frac{2 \pi}{3}\right)$(4) $\left(\frac{\pi}{6}, 0\right)$ [JEE-MAIN 2016]

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Sol. (3)

Q. If the curves $\mathrm{y}^{2}=6 \mathrm{x}, 9 \mathrm{x}^{2}+\mathrm{by}^{2}=16$ intersect each other at right angles, then the value of $\mathrm{b}$ is :(1) $\frac{7}{2}$             (2) 4              (3) $\frac{9}{2}$                 (4) 6 [JEE-MAIN 2016]

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Sol. (3)

• April 15, 2021 at 11:12 pm

Oh bhai 2016 ke aage bhi badho 2021 ke bhi 2 session ho chuke tum itne pichde kyu ho

• December 1, 2020 at 5:19 pm

Op questions

• October 29, 2020 at 10:19 pm

nice

• October 20, 2020 at 5:05 pm

thanx

• October 14, 2020 at 9:49 pm

Good to see all questions here

• October 11, 2020 at 2:39 pm

it is a good platform but no of questions are quite less …

• June 4, 2022 at 2:42 pm

Super

• September 4, 2020 at 3:55 pm

2 nd last question

• July 21, 2020 at 4:25 am

THANKS…FOR ….QUESTIONS

• October 20, 2020 at 3:14 pm

👍👍

• December 1, 2020 at 5:20 pm

Op questions

• July 10, 2020 at 12:08 am

plz can u explain second last question more properly??

• March 4, 2021 at 12:35 pm

U just differentiate given f(x) u will get slope of tangent we known (slopeoftangent*slopeofnormal=-1) from here u get slope of normal they gave x value find y value by sub in eqn of curve then apply normal eqn then sub option one by one u will get answer !

• May 20, 2020 at 4:21 pm

good but need some more problems..

• October 20, 2020 at 3:14 pm

Yes

• February 22, 2021 at 12:54 pm

Good but some more prblms

• February 8, 2022 at 2:36 pm

hi

• May 2, 2020 at 1:46 pm

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• May 2, 2020 at 1:46 pm

well gud can have more stuff and content

• April 28, 2020 at 5:29 pm

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• April 17, 2020 at 3:47 pm

Some more

• March 11, 2022 at 1:02 am

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• April 14, 2020 at 5:51 pm

Thank you