Q.Find the slope of the tangent to the following curves at the respective given points.

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 $4$.