Find the equations of tangent and normal to the curve at the point on it: $2xy + \pi\sin y = 2\pi$ at $\left(1, \dfrac{\pi}{2}\right)$.

Find the equations of tangent and normal to the curve at the point on it: 2xy + πsin y = 2π at (1, (π)/(2)).

Exercise 2.1 · Q4

2xy + \pi\sin y = 2\pi

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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Given $2xy + \pi\sin y = 2\pi$ . Differentiating implicitly:

$2\left(y + x\dfrac{dy}{dx}\right) + \pi\cos y\,\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx}(2x + \pi\cos y) = -2y \Rightarrow \dfrac{dy}{dx} = \dfrac{-2y}{2x+\pi\cos y}$

At $\left(1,\tfrac{\pi}{2}\right)$ : $\cos\left(\tfrac{\pi}{2}\right) = 0$ , so denominator $= 2(1) + \pi(0) = 2$ ; numerator $= -2\left(\tfrac{\pi}{2}\right) = -\pi$ . So $m = -\dfrac{\pi}{2}$ .

Tangent: $y - \dfrac{\pi}{2} = -\dfrac{\pi}{2}(x-1) \Rightarrow 2y - \pi = -\pi x + \pi \Rightarrow \pi x + 2y - 2\pi = 0$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$y = x^2 + 4x + 1$
(a) $x + 10y = 8$
(b) $10x + y = 8$
(c) $10x - y = 8$
(d) $x - 10y = 8$
›Reveal solutionSolution

The tangent at $x=3$ is $10x - y = 8$ .

Given $y = x^2+4x+1$ . The slope is

$\frac{dy}{dx} = 2x+4.$

At $x=3$ : slope $= 2(3)+4 = 10$ , and $y = 9+12+1 = 22$ , so the point is $(3,22)$ .

Equation of the tangent: …

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CBSE 2023Set ANNUAL1 markMCQ
$y^2 = 4ax$
(a) $t$
(b) $\dfrac{2}{t}$
(c) $\dfrac{1}{t}$
(d) None of these
›Reveal solutionSolution

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

For $y^2=4ax$ , the point $(at^2,2at)$ satisfies $x=at^2$ , $y=2at$ .

$\dfrac{dx}{dt}=2at$ , $\dfrac{dy}{dt}=2a$

…

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CBSE 2019Set ANNUAL1 mark
$y = x^3 - x$
›Reveal solutionSolution

The slope of the tangent at a point equals $\dfrac{dy}{dx}$ evaluated at that point.

Given $y = x^3 - x$

$\dfrac{dy}{dx} = 3x^2 - 1$

…

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CBSE 2018Set ANNUAL1 markMCQ
$x = a\cos^3\theta$
(a) $1$
(b) $2$
(c) $-1$
(d) None of these
›Reveal solutionSolution

Find $dy/dx$ for the parametric curve and evaluate at $\theta=\pi/4$ .

$x = a\cos^3\theta \Rightarrow \dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$ .

$y = a\sin^3\theta \Rightarrow \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$ .

…

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CBSE 2017Set ANNUAL1 markMCQ
$y = x^3 - 11x + 5$
(a) $(-2, 0)$
(b) $(3, 7)$
(c) $(0, 2)$
(d) $(2, -9)$
›Reveal solutionSolution

Matching the curve's slope to the tangent's slope $1$ gives the point $(2,-9)$ .

$y=x^3-11x+5\Rightarrow \dfrac{dy}{dx}=3x^2-11$ . The tangent line $y=x-11$ has slope $1$ , so $3x^2-11=1\Rightarrow x^2=4\Rightarrow x=\pm2$ .

…

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Q.Find the equations of tangent and normal to the curve at the point on it: $2xy + \pi\sin y = 2\pi$ at $\left(1, \dfrac{\pi}{2}\right)$ .

Concept understanding — Tangent and Normal to a Curve

Q.Find the equations of tangent and normal to the curve at the point on it: 2xy+πsin⁡y=2π2xy + \pi\sin y = 2\pi2xy+πsiny=2π at (1,π2)\left(1, \dfrac{\pi}{2}\right)(1,2π​).

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Q.Find the equations of tangent and normal to the curve at the point on it: $2xy + \pi\sin y = 2\pi$ at $\left(1, \dfrac{\pi}{2}\right)$ .

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