Q.The equation of the normal to the parabola $y^2 = 8x$ at its origin is ________.

Showing the 12 most recent of 67 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\tan(e^{3x}) = \cot(e^{2y})$, then at $x = 0$, $\dfrac{dy}{dx} =$ (A) $\dfrac{3}{2-\pi}$ (B) $\dfrac{2-\pi}{3}$ (C) $\dfrac{3}{\pi-2}$ (D) $\dfrac{\pi-2}{3}$

   ›Reveal solutionSolution

   Rewrite $\cot$ as a shifted $\tan$ to turn the equation into an algebraic (exponential) relation between $x$ and $y$, then implicitly differentiate and evaluate at $x=0$. Answer: $\dfrac{3}{2-\pi}$.

   Concept and Intuition

   $\tan\theta_1=\tan\theta_2$ implies $\theta_1=\theta_2+n\pi$ for integer $n$; taking $n=0$ (the principal relation intended here) converts the trig equation into a clean equation between the exponential expressions, which we can differentiate implicitly.

   Step-by-Step Solution

   1. Use the identity $\cot\theta = \tan\left(\dfrac{\pi}{2}-\theta\right)$ with $\theta=e^{2y}$: $\cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$.
   2. Given $\tan(e^{3x}) = \cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$, equate arguments (principal branch): $e^{3x} = \dfrac{\pi}{2}-e^{2y}$.
   3. Rearrange: $e^{3x}+e^{2y} = \dfrac{\pi}{2}$.
   4. Differentiate both sides w.r.t. $x$: $3e^{3x} + 2e^{2y}\dfrac{dy}{dx} = 0$.
   5. Solve: $\dfrac{dy}{dx} = -\dfrac{3e^{3x}}{2e^{2y}}$.
   6. At $x=0$: $e^{3x}=e^0=1$. From step 3, $1+e^{2y}=\dfrac{\pi}{2}\Rightarrow e^{2y}=\dfrac{\pi}{2}-1=\dfrac{\pi-2}{2}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\mathrm{Tan}^{-1}x^2 + \mathrm{Tan}^{-1}y^2 = \dfrac{\pi}{2}$, then $\left(\dfrac{dy}{dx}\right)_{(-1,2)} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   Reducing to $y^2=x^{-2}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$, which at $(-1,2)$ equals $\dfrac12$.

   Concept and Intuition

   If $\mathrm{Tan}^{-1}a+\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}$ with $a,b>0$, then $\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}-\mathrm{Tan}^{-1}a=\mathrm{Cot}^{-1}a$, so $b=\dfrac{1}{a}$. Applying this to $a=x^2,\ b=y^2$ collapses the relation into an algebraic one.

   Step-by-Step Solution

   1. From $\mathrm{Tan}^{-1}x^2+\mathrm{Tan}^{-1}y^2=\dfrac{\pi}{2}$ we get $y^2=\dfrac{1}{x^2}=x^{-2}$.
   2. Differentiate: $2y\,\dfrac{dy}{dx}=-2x^{-3}$.
   3. Hence $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the vertical angle of a cone is $60^\circ$ and the rate of change of its total surface area is $2\sqrt{3}\ cm^2/sec$, then the rate of change of its volume (in $cm^3/sec$) when its radius is $5\ cm$, is (A) $15$ (B) $10$ (C) $5$ (D) $9$

   ›Reveal solutionSolution

   A related-rates problem: using the $60°$ vertical angle to fix $h$ and slant height $l$ in terms of $r$, then chaining $\dfrac{dS}{dt}\to\dfrac{dr}{dt}\to\dfrac{dV}{dt}$ gives $5\ \text{cm}^3/\text{sec}$.

   Concept and Intuition

   When a cone's vertical (apex) angle is fixed, its shape stays similar as it grows — radius and height stay in a fixed ratio determined by the semi-vertical angle. This lets us express both surface area and volume purely in terms of $r$, so a single related-rates chain (through $dr/dt$) connects the given rate of surface-area change to the unknown rate of volume change.

   Step-by-Step Solution

   1. Vertical angle $=60°$, so semi-vertical angle $\alpha=30°$. In the cone's cross-section, $\tan\alpha = r/h$, so $r = h\tan30° = h/\sqrt3$, i.e. $h = r\sqrt3$.
   2. Slant height: $l = \sqrt{r^2+h^2} = \sqrt{r^2+3r^2} = \sqrt{4r^2} = 2r$.
   3. Total surface area: $S = \pi r^2 + \pi r l = \pi r^2 + \pi r(2r) = 3\pi r^2$.
   4. Differentiate: $\dfrac{dS}{dt} = 6\pi r\dfrac{dr}{dt}$.
   5. Given $\dfrac{dS}{dt}=2\sqrt3$ and $r=5$: $2\sqrt3 = 6\pi(5)\dfrac{dr}{dt} = 30\pi\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{2\sqrt3}{30\pi} = \dfrac{\sqrt3}{15\pi}$.
   6. Volume: $V = \dfrac13\pi r^2 h = \dfrac13\pi r^2(r\sqrt3) = \dfrac{\pi r^3}{\sqrt3}$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The surface area of a sphere is $49\pi\ sq.cm$. If it is increased by $0.016\ sq.cm$, then the approximate increase in its volume (in $c.c.$) is (A) $0.07$ (B) $0.04$ (C) $0.032$ (D) $0.028$

   ›Reveal solutionSolution

   Using differentials to connect a small change in surface area to the corresponding small change in volume, via the shared variable $r$, gives an approximate volume increase of $0.028$ c.c.

   Concept and Intuition

   When a small change in one geometric quantity (surface area) causes a small change in another (volume), and both depend on a common variable ($r$), we can relate their differentials directly: $dS = 8\pi r\,dr$ and $dV = 4\pi r^2\,dr$ share the same $dr$, so we solve for $dr$ from the given $dS$ and substitute into the $dV$ formula — this is the standard "approximate change" technique using derivatives.

   Step-by-Step Solution

   1. Surface area of sphere: $S=4\pi r^2 = 49\pi \Rightarrow r^2 = \dfrac{49}{4} \Rightarrow r = \dfrac72 = 3.5\ \text{cm}$.
   2. Differentiate $S=4\pi r^2$: $dS = 8\pi r\, dr$.
   3. Given $dS = 0.016$, and $r=3.5$: $dr = \dfrac{0.016}{8\pi(3.5)} = \dfrac{0.016}{28\pi}$.
   4. Volume: $V = \dfrac43\pi r^3$. Differentiate: $dV = 4\pi r^2\, dr$.
   5. $r^2 = 12.25$, so $dV = 4\pi(12.25)\,dr = 49\pi\, dr$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the normal drawn to the curve $y^4 = 16x^3$ at the point of intersection of this curve and the line $y = 2$ meets the X and Y axes at $A$ and $B$ respectively, then $OA + 3\,OB =$ (A) $6$ (B) $8$ (C) $16$ (D) $12$

   ›Reveal solutionSolution

   Finding the intersection point, then the normal line's slope and intercepts, gives $OA+3\,OB=12$.

   Concept and Intuition

   The normal to a curve at a point is the line perpendicular to the tangent there. We find the point of intersection with $y=2$, use implicit differentiation to get the tangent slope, take its negative reciprocal for the normal slope, then find where that normal line crosses each axis (its intercepts) to compute $OA$ and $OB$ (the distances from the origin to those intercepts).

   Step-by-Step Solution

   1. Find the intersection: substitute $y=2$ into $y^4=16x^3$: $16 = 16x^3 \Rightarrow x^3=1 \Rightarrow x=1$. So the point is $(1,2)$.
   2. Differentiate implicitly: $4y^3\dfrac{dy}{dx} = 48x^2 \Rightarrow \dfrac{dy}{dx} = \dfrac{12x^2}{y^3}$.
   3. At $(1,2)$: $\dfrac{dy}{dx} = \dfrac{12(1)}{8} = \dfrac32$. This is the tangent's slope.
   4. Normal's slope is the negative reciprocal: $m_{normal} = -\dfrac{2}{3}$.
   5. Normal line equation: $y - 2 = -\dfrac23(x-1)$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2=t+\dfrac{1}{t}$ and $x^4+y^4=t^2+\dfrac{1}{t^2}$, then $x^3 y\dfrac{dy}{dx}=$ (A) -1 (B) 0 (C) 1 (D) 2

   ›Reveal solutionSolution

   The two given relations force $x^2y^2=1$ (i.e. $xy$ is constant), from which $x^3y\,dy/dx=-1$.

   Concept and Intuition

   Rather than solving for $x,y$ in terms of $t$ explicitly, combine the two given equations algebraically (square the first, subtract the second) to eliminate $t$ entirely and land on a simple constant-product relation between $x$ and $y$.

   Step-by-Step Solution

   1. Square the first relation: $(x^2+y^2)^2=\left(t+\dfrac1t\right)^2=t^2+2+\dfrac{1}{t^2}$, i.e.

   $x^4+2x^2y^2+y^4=t^2+2+\frac{1}{t^2}$

   2. The second given relation is $x^4+y^4=t^2+\dfrac{1}{t^2}$.
   3. Subtract: $2x^2y^2 = \left(t^2+2+\dfrac1{t^2}\right)-\left(t^2+\dfrac1{t^2}\right)=2$, so $x^2y^2=1$.
   4. This means $xy=\pm1$, a constant independent of $t$. Differentiate $xy=\text{const}$ implicitly: …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2+\sin y=4$, then the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,0)$ is (A) -34 (B) -32 (C) 34 (D) 32

   ›Reveal solutionSolution

   Implicit differentiation twice on $x^2+y^2+\sin y=4$ gives $y''=-34$ at $(-2,0)$.

   Concept and Intuition

   For an implicitly defined curve, differentiate the whole equation with respect to $x$ once to get $y'$ in terms of $x,y$, then differentiate that resulting equation again (product/chain rule carefully) to isolate $y''$, finally substituting the numeric point.

   Step-by-Step Solution

   1. Differentiate $x^2+y^2+\sin y=4$ w.r.t. $x$:

   $2x+2y\,y'+\cos y\,y'=0 \Rightarrow y'(2y+\cos y)=-2x \Rightarrow y'=\frac{-2x}{2y+\cos y}$

   2. At $(-2,0)$: $y'=\dfrac{-2(-2)}{2(0)+\cos0}=\dfrac{4}{1}=4$.
   3. Differentiate the equation $2x+2y y'+\cos y\, y'=0$ again w.r.t. $x$:

   $2+2(y')^2+2y\,y''-\sin y\,(y')^2+\cos y\,y''=0$

   $2+\left[2-\sin y\right](y')^2+(2y+\cos y)y''=0$

   4. At the point: $y=0$, $y'=4$, $\sin y=0$, $\cos y=1$, so $2y+\cos y=1$: …

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8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Let the normal drawn at a point $P$ on the curve $y^2-3x^2+y+10=0$ intersect the Y-axis at $\left(0,\dfrac{3}{2}\right)$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|=$ (A) 2 (B) 3 (C) 4 (D) 6

   ›Reveal solutionSolution

   Implicit differentiation gives the tangent slope in terms of $(x_0,y_0)$; using that the normal meets the $Y$-axis at $(0,3/2)$ pins down $y_0=1$, then the curve equation gives $x_0=\pm2$, so $|m|=4$.

   Concept and Intuition

   When a curve is given implicitly, differentiate term-by-term treating $y$ as a function of $x$ to get $dy/dx$ symbolically in terms of $x,y$. The normal line at a point has slope $-1/m$ where $m$ is the tangent slope; using where that normal crosses a known axis gives an equation linking $x_0,y_0$, which combined with the original curve equation pins down the point.

   Step-by-Step Solution

   1. Differentiate $y^2-3x^2+y+10=0$: $2y\,y' - 6x + y' = 0 \Rightarrow y'(2y+1)=6x \Rightarrow y' = \dfrac{6x}{2y+1}$.
   2. At $P=(x_0,y_0)$, tangent slope $m = \dfrac{6x_0}{2y_0+1}$, so normal slope $= -\dfrac{2y_0+1}{6x_0}$.
   3. Equation of the normal at $P$: $y - y_0 = -\dfrac{1}{m}(x-x_0)$. At $x=0$, $y=\dfrac{3}{2}$: $\dfrac{3}{2}-y_0 = -\dfrac{1}{m}(0-x_0) = \dfrac{x_0}{m}$.
   4. Since $m = \dfrac{6x_0}{2y_0+1}$, we get $\dfrac{x_0}{m} = \dfrac{2y_0+1}{6}$ (the $x_0$ cancels). …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If a cylindrical tank of radius 3 m is filled with water at the rate of $\dfrac{3}{2}$ $m^3$/sec, then the rate of change of its water level in (m/sec) is (A) $\dfrac{1}{3\pi}$ (B) $\dfrac{1}{2\pi}$ (C) $\dfrac{1}{\pi}$ (D) $\dfrac{1}{6\pi}$

   ›Reveal solutionSolution

   This tests related rates for a cylinder of fixed radius; the water level rises at $\dfrac{1}{6\pi}$ m/s.

   Concept and Intuition

   Since the radius doesn't change as the tank fills, the volume $V=\pi r^2h$ is a function of $h$ alone (with $r$ a constant), so differentiating with respect to time directly links $dV/dt$ to $dh/dt$ through the constant cross-sectional area $\pi r^2$.

   Step-by-Step Solution

   1. $V = \pi r^2 h$, with $r=3$ constant, so $V=9\pi h$.
   2. Differentiate w.r.t. time: $\dfrac{dV}{dt} = 9\pi\dfrac{dh}{dt}$. …

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10. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.The length of the tangent drawn at the point $P(1, 3\sqrt{3})$ on the curve $x^{2/3} + y^{2/3} = 4$ is (A) 4 (B) 6 (C) 12 (D) 8

    ›Reveal solutionSolution

    This tests the "length of tangent" formula (segment of the tangent line between the point of contact and the $x$-axis) applied to a point on an astroid. Answer: length $=6$.

    Concept and Intuition

    For a curve $y=f(x)$, the tangent line at $(x_1,y_1)$ with slope $m$ meets the $x$-axis at $\left(x_1-\dfrac{y_1}{m},0\right)$. The distance from $(x_1,y_1)$ to that point is called the length of the tangent, and it works out to

    $L=|y_1|\frac{\sqrt{1+m^2}}{|m|}.$

    So we just need $y_1$ and the slope $m=\left.\dfrac{dy}{dx}\right|_P$ from the implicit equation of the astroid.

    Step-by-Step Solution

    1. The curve is $x^{2/3}+y^{2/3}=4$. Check $P(1,3\sqrt3)$: $1^{2/3}=1$ and $(3\sqrt3)^{2/3}=(3^{3/2})^{2/3}=3$, so $1+3=4$ — $P$ lies on the curve.
    2. Differentiate implicitly: $\dfrac{2}{3}x^{-1/3}+\dfrac{2}{3}y^{-1/3}\dfrac{dy}{dx}=0\Rightarrow \dfrac{dy}{dx}=-\left(\dfrac{y}{x}\right)^{1/3}$. …

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11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.If $x^2y - xy^2 + x^3 - y^3 = 0$, then $\dfrac{dy}{dx}$ at the point $(1, 1)$ is (A) $1$ (B) $0$ (C) $-1$ (D) Does not exist

    ›Reveal solutionSolution

    Implicit differentiation of a symmetric cubic curve, evaluated at the point $(1,1)$.

    Concept and Intuition

    When a curve is given implicitly (not solved for $y$), differentiate every term with respect to $x$, treating $y$ as a function of $x$ and applying the product rule wherever $x$ and $y$ appear together. Collecting all the $y'$ terms on one side isolates the slope as a ratio of two expressions in $x,y$.

    Step-by-Step Solution

    1. Differentiate term by term: $\dfrac{d}{dx}(x^2y)=2xy+x^2y'$; $\dfrac{d}{dx}(xy^2)=y^2+2xyy'$; $\dfrac{d}{dx}(x^3)=3x^2$; $\dfrac{d}{dx}(y^3)=3y^2y'$.
    2. So $2xy+x^2y'-y^2-2xyy'+3x^2-3y^2y'=0$.
    3. Collect $y'$ terms: $y'(x^2-2xy-3y^2)=-(2xy-y^2+3x^2)$, i.e. $y'=\dfrac{-(2xy-y^2+3x^2)}{x^2-2xy-3y^2}=\dfrac{y^2-2xy-3x^2}{x^2-2xy-3y^2}$. …

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12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.If the rate of increase in the surface area of a cube is $6\ sq.cm./sec$, then the rate of increase in its volume (in $c.c./sec$), when the length of its edge is $12\ cm$, is (A) $6$ (B) $12$ (C) $18$ (D) $9$

    ›Reveal solutionSolution

    A related-rates problem: convert the given rate of change of surface area into the rate of change of the edge length, then into the rate of change of volume.

    Concept and Intuition

    Both surface area and volume of a cube are functions of the single variable $a$ (the edge length), so differentiating each with respect to time and using the chain rule links their rates through $da/dt$ — the one quantity actually changing independently.

    Step-by-Step Solution

    1. Surface area: $S=6a^2\Rightarrow\dfrac{dS}{dt}=12a\dfrac{da}{dt}$.
    2. Given $\dfrac{dS}{dt}=6$: $12a\dfrac{da}{dt}=6\Rightarrow\dfrac{da}{dt}=\dfrac{1}{2a}$.
    3. Volume: $V=a^3\Rightarrow\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}$. …

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