Find the equations of tangent and normal to the curve at the point on it: $x=\sqrt t,\ y = t - \dfrac{1}{\sqrt t}$ at $t=4$.

Find the equations of tangent and normal to the cu – Class 12 Mathematics

Exercise 2.1 · Q7

x=\sqrt t,\ y = t - \dfrac{1}{\sqrt t}

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Given $x=\sqrt t=t^{1/2}$ , $y = t - t^{-1/2}$ . $\dfrac{dx}{dt} = \dfrac{1}{2\sqrt t}$ . $\dfrac{dy}{dt} = 1 + \dfrac12 t^{-3/2}$ .

At $t=4$ : $\dfrac{dx}{dt}=\dfrac{1}{2(2)}=\dfrac14$ . $\dfrac{dy}{dt}=1+\dfrac12\cdot\dfrac{1}{8}=1+\dfrac{1}{16}=\dfrac{17}{16}$ .

$\dfrac{dy}{dx} = \dfrac{17/16}{1/4} = \dfrac{17}{16}\times4 = \dfrac{17}{4}$ .

Point: $x=\sqrt4=2$ , $y=4-\dfrac{1}{2}=\dfrac72$ . …

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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: $x=\sqrt t,\ y = t - \dfrac{1}{\sqrt t}$ at $t=4$ .

Concept understanding — Tangent and Normal to a Curve

Q.Find the equations of tangent and normal to the curve at the point on it: x=t, y=t−1tx=\sqrt t,\ y = t - \dfrac{1}{\sqrt t}x=t​, y=t−t​1​ at t=4t=4t=4.

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Q.Find the equations of tangent and normal to the curve at the point on it: $x=\sqrt t,\ y = t - \dfrac{1}{\sqrt t}$ at $t=4$ .

Unlock everything free for 14 days