Q.Find the equations of the tangent and the normal to the curve $y = x^3$ at the point $(1, 1)$.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set SEM31 markMCQ
   Q.If the straight line $y = x$ and the curve $xy = k^2$ cut at a right angle, then ($k$ is a real constant)

   (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

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

   Product of slopes $= 1\times(-1) = -1$, so the curves cut orthogonally. This holds for every real $k$ (the intersection point $x=\pm k$ exists for all real $k$).

   ✓Final answer

   The curves cut at a right angle for all real $k$, i.e. $-\infty<k<\infty$ — option (c).

2. CBSE 2026Set SEM31 markMCQ
   Q.If the straight line $lx - my + n = 0$ touches the parabola $y^2 = 4ax$ then

   (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:

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

   A line $y=Mx+c$ is tangent to $y^2=4ax$ iff

   $c = \frac{a}{M}.$

   Substitute:

   $\frac{n}{m} = \frac{a}{\,l/m\,} = \frac{am}{l}.$

   Cross-multiplying: $nl = am^2$, i.e. $am^2 = nl$.

   ✓Final answer

   Condition of tangency: $am^2 = nl$ — option (a).

3. CBSE 2024Set EX1 markMCQ
   Q.At which point is the slope of the line $y = x + 1$ equal to the slope of the curve $y^2 = 4x$?

   (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:

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

   Set equal to the line's slope: $\dfrac{2}{y}=1\Rightarrow y=2$.

   Substitute into the parabola: $y^2=4x\Rightarrow 4=4x\Rightarrow x=1$. The point is $(1,2)$, which indeed satisfies $y^2=4x$.

   ✓Final answer

   (a) $(1,2)$.

4. CBSE 2024Set ANNUAL1 mark
   Q.Write the point where the tangent to the curve $y^2 - x^2 + 2x - 1 = 0$ is parallel to the $x$-axis.

   ›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 $x$:

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

   The tangent is parallel to the $x$-axis where $\dfrac{dy}{dx}=0$, i.e. $x=1$.

   Substituting $x=1$ into the curve: $y^2 - 1 + 2 - 1 = 0 \Rightarrow y^2 = 0 \Rightarrow y = 0$.

   ✓Final answer

   The point is $(1, 0)$.

5. CBSE 2024Set ANNUAL1 mark
   Q.Find the slope of tangent to the curve $2y = 3 - x^3$ at the point (1, 1).

   ›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 $x=1$:

   $\dfrac{dy}{dx}\Big|_{(1,1)} = -\dfrac{3(1)^2}{2} = -\dfrac{3}{2}$

   ✓Final answer

   Slope of the tangent $= -\dfrac{3}{2}$.

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

   Step 3. This is the slope at any parameter $t$, in particular at $(ct,c/t)$.

   ✓Final answer

   Slope $=-\dfrac{1}{t^2}$ (option a).

7. CBSE 2023Set ANNUAL1 markMCQ
   Q.Angle between the curves $y^2=x$ and $x^2=y$ at the origin is :

   (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.

   1. 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 $x=0$.
   2. Curve $x^2=y$: $\dfrac{dy}{dx}=2x$. At the origin $x=0$, so $\dfrac{dy}{dx}=0$ — the tangent is the horizontal line $y=0$.
   3. The angle between the curves at the origin is the angle between their tangent lines, i.e. between $x=0$ and $y=0$, which is $\dfrac{\pi}{2}$ (a right angle).

   ✓Final answer

   (a) $\dfrac{\pi}{2}$

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.The abscissa of the point on the curve $f(x)=\sqrt{8-2x}$ at which the slope of the tangent is $-0.25$ ?

   (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 $x=-4$.

   1. $f(x)=\sqrt{8-2x}=(8-2x)^{1/2}$. By the chain rule, $f'(x)=\dfrac12(8-2x)^{-1/2}\cdot(-2)=\dfrac{-1}{\sqrt{8-2x}}$.
   2. Set $f'(x)=-0.25=-\dfrac14$: $\dfrac{-1}{\sqrt{8-2x}}=-\dfrac14\Rightarrow\sqrt{8-2x}=4$.
   3. Square both sides: $8-2x=16\Rightarrow -2x=8\Rightarrow x=-4$.
   4. Check: at $x=-4$, $8-2(-4)=16$, and $f(x)$ is defined ($16\ge0$). Valid.

   ✓Final answer

   (d) $-4$

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

   At $x=1$: $\dfrac{dy}{dx} = 4-18+26-10 = 2$. This is the slope $m$ of the tangent.

   Using point-slope form at $(1,3)$:

   $y - 3 = 2(x-1) \implies y = 2x + 1 \implies 2x - y + 1 = 0.$

   ✓Final answer

   2x − y + 1 = 0 — option (c).

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.What is the slope of the tangent to the curve $y = \ln x\ (x>0)$ at $x=1$?

    (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.

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

    At $x=1$: slope $=\dfrac11=1$.

    ✓Final answer

    (b) $1$.

11. CBSE 2022Set ANNUAL1 mark
    Q.For $x =$ ____, the tangent to the curve $y=\cos x,\ 0\le x\le\pi$, is parallel with $Y$-axis. Choices given: $[\frac{\pi}{4},\ \frac{\pi}{3},\ \frac{\pi}{2},\ \pi]$

    ›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.

    Treating the intended question as the standard textbook one — tangent parallel to the $X$-axis — we need $\dfrac{dy}{dx}=0$: $-\sin x=0\Rightarrow x=0$ or $x=\pi$ within $[0,\pi]$. Of these, only $x=\pi$ appears among the given choices $\left[\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2},\pi\right]$.

    ✓Final answer

    $x=\pi$ — the point where the tangent is horizontal (parallel to the $X$-axis); the stem's wording "$Y$-axis" is read as the standard textbook version. Verified by two experienced subject lecturers.

12. 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.

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

    At the point $(2,5)$: $\dfrac{dy}{dx}\Big|_{x=2} = 2(2) = 4$.

    ✓Final answer

    Slope of the tangent at $(2,5)$ is $4$.