Find the equations of the tangent and normal at $\theta=\dfrac{\pi}{2}$ to the curve $x=a(\theta+\sin\theta)$, $y=a(1+\cos\theta)$.

> [!TLDR] > Find $dy/dx$ in terms of the parameter $\theta$, evaluate at $\theta=\pi/2$, then write the point-slope equations for the tangent and the normal. > > 1. **Differentiate $x$ and $y$ with respect to $\theta$.** > > $x=a(\theta+\sin\theta) \Rightarrow \dfrac{dx}{d\theta}=a(1+\cos\theta)$ > > $y=a(1+\cos\theta) \Rightarrow \dfrac{dy}{d\theta}=-a\sin\theta$ > > 2. **Form $\dfrac{dy}{dx}$.** > > $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{-a\sin\theta}{a(1+\cos\theta)}=\dfrac{-\sin\theta}{1+\cos\theta}$ > > 3. **Evaluate the point at $\theta=\dfrac{\pi}{2}$.** Here $\sin\dfrac{\pi}{2}=1$, $\cos\dfrac{\pi}{2}=0$. > > $x=a\left(\dfrac{\pi}{2}+1\right)$, $\quad y=a(1+0)=a$ > > 4. **Evaluate the slope at $\theta=\dfrac{\pi}{2}$.** > > $\dfrac{dy}{dx}=\dfrac{-1}{1+0}=-1$ > > 5. **Equation of the tangent** (slope $m=-1$ through $\left(a\left(\dfrac{\pi}{2}+1\right),\,a\right)$): > > $y-a=-1\left[x-a\left(\dfrac{\pi}{2}+1\right)\right]$ > > $y-a=-x+a\left(\dfrac{\pi}{2}+1\right)$ > …

Find the equations of the tangent and normal at θ=(π)/(2) to the curve x=a(θ+sinθ), y=a(1+cosθ).

Question 93 of 148

\theta=\dfrac{\pi}{2}

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★

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Find $dy/dx$ in terms of the parameter $\theta$ , evaluate at $\theta=\pi/2$ , then write the point-slope equations for the tangent and the normal.

Differentiate $x$ and $y$ with respect to $\theta$ .

$x=a(\theta+\sin\theta) \Rightarrow \dfrac{dx}{d\theta}=a(1+\cos\theta)$

$y=a(1+\cos\theta) \Rightarrow \dfrac{dy}{d\theta}=-a\sin\theta$
Form $\dfrac{dy}{dx}$ .

$\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{-a\sin\theta}{a(1+\cos\theta)}=\dfrac{-\sin\theta}{1+\cos\theta}$
Evaluate the point at $\theta=\dfrac{\pi}{2}$ . Here $\sin\dfrac{\pi}{2}=1$ , $\cos\dfrac{\pi}{2}=0$ .

$x=a\left(\dfrac{\pi}{2}+1\right)$ , $\quad y=a(1+0)=a$
Evaluate the slope at $\theta=\dfrac{\pi}{2}$ .

$\dfrac{dy}{dx}=\dfrac{-1}{1+0}=-1$
Equation of the tangent (slope $m=-1$ through $\left(a\left(\dfrac{\pi}{2}+1\right),\,a\right)$ ):

$y-a=-1\left[x-a\left(\dfrac{\pi}{2}+1\right)\right]$

$y-a=-x+a\left(\dfrac{\pi}{2}+1\right)$

…

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CBSE 2026Set SEM31 markMCQ
$y = x$
(a) $k = 0$
(b) $k = \pm 1$
(c) $-\infty < k < \infty$
(d) $0 \le k < \infty$
›Reveal solutionSolution

The line has slope $1$ ; the curve's slope at any point of $y=x$ works out to $-1$ , so they always meet at right angles for any real $k$ .

The tangent-slope/orthogonality condition is a CBSE/NCERT Class 12 application of derivatives topic.

The line $y=x$ has slope $1$ .

For the curve $xy=k^2$ , differentiate implicitly: $y + x\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}$ .

At an intersection point lying on $y=x$ we have $y=x$ , so $x\cdot x = k^2 \Rightarrow x^2=k^2$ , and the curve's slope is

$\frac{dy}{dx} = -\frac{y}{x} = -\frac{x}{x} = -1.$

…

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CBSE 2026Set SEM31 markMCQ
$lx - my + n = 0$
(a) $am^2 = nl$
(b) $an^2 = ml$
(c) $al^2 = mn$
(d) $mn = al$
›Reveal solutionSolution

Apply the standard tangency condition $c=\dfrac{a}{M}$ for a line $y=Mx+c$ touching $y^2=4ax$ ; it yields $am^2=nl$ .

The tangent condition to a parabola is a coordinate-geometry / conic-sections result (aligned with the NCERT/CBSE coordinate-geometry stream), included here for completeness even though the chapter is not in the given menu.

Write the line $lx-my+n=0$ in slope form:

$y = \frac{l}{m}x + \frac{n}{m}, \quad \text{so } M=\frac{l}{m},\ c=\frac{n}{m}.$

…

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CBSE 2024Set EX1 markMCQ
$y = x + 1$
(a) $(1, 2)$
(b) $(2, 1)$
(c) $(1, -2)$
(d) $(-1, 2)$
›Reveal solutionSolution

The line's slope is $1$ . Differentiating $y^2=4x$ gives slope $\dfrac{2}{y}$ ; set $\dfrac{2}{y}=1\Rightarrow y=2$ , then $x=1$ . Point $(1,2)$ , option (a).

Concept. The slope of a curve at a point is $\dfrac{dy}{dx}$ there. We need the point on the parabola where this equals the constant slope of the given line.

Line. $y=x+1\Rightarrow \dfrac{dy}{dx}=1$ .

Curve. Differentiate $y^2=4x$ implicitly:

$2y\,\frac{dy}{dx}=4 \;\Rightarrow\; \frac{dy}{dx}=\frac{2}{y}.$ …

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CBSE 2024Set ANNUAL1 mark
$y^2 - x^2 + 2x - 1 = 0$
›Reveal solutionSolution

Differentiating implicitly and setting $\dfrac{dy}{dx}=0$ gives $x=1$ , and substituting back into the curve gives $y=0$ .

Differentiate $y^2-x^2+2x-1=0$ implicitly with respect to $x$ :

$2y\dfrac{dy}{dx} - 2x + 2 = 0 \implies \dfrac{dy}{dx} = \dfrac{2x-2}{2y} = \dfrac{x-1}{y}$

…

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CBSE 2024Set ANNUAL1 mark
$2y = 3 - x^3$
›Reveal solutionSolution

Differentiate implicitly to get $\dfrac{dy}{dx}$ , then substitute $x=1$ .

Given curve: $2y = 3-x^3$ .

Differentiate both sides with respect to $x$ :

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

At the point $(1,1)$ , substitute $x=1$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Slope of tangent of xy = c² at (ct, c/t) is$
(a) -1/t²
(b) 1/t²
(c) -1/t
(d) -1/t³
›Reveal solutionSolution

For a parametric curve, the slope of the tangent is $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

Step 1. $x=ct,\ y=c/t$ . Then $\dfrac{dx}{dt}=c$ and $\dfrac{dy}{dt}=-\dfrac{c}{t^2}$ .

Step 2. $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-c/t^2}{c} = -\dfrac{1}{t^2}$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$y^2=x$
(a) $\dfrac{\pi}{2}$
(b) $\tan^{-1}\left(\dfrac34\right)$
(c) $\dfrac{\pi}{4}$
(d) $\tan^{-1}\left(\dfrac43\right)$
›Reveal solutionSolution
Finding the tangent line to each curve at the origin shows one is vertical and the other horizontal, so they meet at a right angle.

Curve $y^2=x$ : differentiate implicitly, $2y\,\dfrac{dy}{dx}=1\Rightarrow\dfrac{dy}{dx}=\dfrac{1}{2y}$ . At the origin $y=0$ , this is undefined — equivalently $\dfrac{dx}{dy}=2y=0$ at $y=0$ , so the tangent is the vertical line $x=0$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$f(x)=\sqrt{8-2x}$
(a) $-2$
(b) $-8$
(c) $0$
(d) $-4$
›Reveal solutionSolution
Differentiating $f(x)=\sqrt{8-2x}$ and setting the slope to $-0.25$ pins down $x=-4$ .

$f(x)=\sqrt{8-2x}=(8-2x)^{1/2}$ . By the chain rule, $f'(x)=\dfrac12(8-2x)^{-1/2}\cdot(-2)=\dfrac{-1}{\sqrt{8-2x}}$ .

Set $f'(x)=-0.25=-\dfrac14$ : $\dfrac{-1}{\sqrt{8-2x}}=-\dfrac14\Rightarrow\sqrt{8-2x}=4$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The equation of the tangent to the curve y = x⁴ - 6x³ + 13x² - 10x + 5 at the point (1, 3) is -$
(a) 2y + x + 1 = 0
(b) −2x + y + 1 = 0
(c) 2x − y + 1 = 0
(d) 2y − x + 1 = 0
›Reveal solutionSolution

Find $dy/dx$ at $x=1$ to get the slope, then use the point-slope form of a line through $(1,3)$ .

Given $y = x^4 - 6x^3 + 13x^2 - 10x + 5$ . Check the point: at $x=1$ , $y = 1-6+13-10+5 = 3$ ✓, confirming $(1,3)$ lies on the curve.

Differentiate: $\dfrac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$y = \ln x\ (x>0)$
(a) $-1$
(b) $1$
(c) $0$
(d) $2$
›Reveal solutionSolution

The slope of the tangent at a point is the value of $dy/dx$ at that point.

$y=\ln x \Rightarrow \dfrac{dy}{dx}=\dfrac1x$ . …

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CBSE 2022Set ANNUAL1 mark
$x =$
›Reveal solutionSolution

A tangent parallel to the $X$ -axis occurs where the slope $dy/dx=0$ ; for $y=\cos x$ on $[0,\pi]$ that happens at $x=0$ and $x=\pi$ .

$y=\cos x \Rightarrow \dfrac{dy}{dx}=-\sin x$ .

Since $-\sin x$ is finite for every $x$ , this curve never has an actual vertical tangent (parallel to the $Y$ -axis) on $[0,\pi]$ — so taken completely literally, the question as worded has no solution among the given choices.

…

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CBSE 2022Set ANNUAL1 mark
$Q. Slope of the tangent to the curve y = x² + 1 at the point (2, 5) is ....$
›Reveal solutionSolution

The slope of the tangent at a point is dy/dx evaluated there.

$y = x^2+1 \Rightarrow \dfrac{dy}{dx} = 2x$ .

…

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Q.Find the equations of the tangent and normal at $\theta=\dfrac{\pi}{2}$ to the curve $x=a(\theta+\sin\theta)$ , $y=a(1+\cos\theta)$ .

Concept understanding — Tangents and Normals (Lengths)

Tangents and Normals to a Curve

Q.Find the equations of the tangent and normal at θ=π2\theta=\dfrac{\pi}{2}θ=2π​ to the curve x=a(θ+sin⁡θ)x=a(\theta+\sin\theta)x=a(θ+sinθ), y=a(1+cos⁡θ)y=a(1+\cos\theta)y=a(1+cosθ).

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Q.Find the equations of the tangent and normal at $\theta=\dfrac{\pi}{2}$ to the curve $x=a(\theta+\sin\theta)$ , $y=a(1+\cos\theta)$ .

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