Q.Find $\frac{dy}{dx}$, if $x^{2/3} + y^{2/3} = a^{2/3}$.

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The length of the tangent drawn at the point $p\left(\dfrac{\pi}{4}\right)$ on the curve $x^{2/3}+y^{2/3}=2^{2/3}$ is (A) $\dfrac{2}{3}$ (B) 1 (C) $\dfrac{4}{3}$ (D) 2

   ›Reveal solutionSolution

   A length-of-tangent computation on the astroid at its symmetric point $\theta=\pi/4$; the length comes out to exactly $1$.

   Concept and Intuition

   The astroid $x^{2/3}+y^{2/3}=a^{2/3}$ can be parametrized as $x=a\cos^3\theta,\ y=a\sin^3\theta$, which automatically satisfies the equation. "The point $P(\pi/4)$" refers to this parameter value. The length of the tangent line segment (from the point of tangency to where it meets the $x$-axis) is given by the standard formula $\left|\dfrac{y\sqrt{1+y'^2}}{y'}\right|$.

   Step-by-Step Solution

   1. Here $a=2$ (since $2^{2/3}=a^{2/3}$), so $x=2\cos^3\theta$, $y=2\sin^3\theta$.
   2. At $\theta=\pi/4$: $\cos(\pi/4)=\sin(\pi/4)=1/\sqrt2$, so $x=y=2\left(\dfrac{1}{\sqrt2}\right)^3=2\cdot\dfrac{1}{2\sqrt2}=\dfrac{1}{\sqrt2}$.
   3. Differentiate $x^{2/3}+y^{2/3}=a^{2/3}$ implicitly: $\dfrac23x^{-1/3}+\dfrac23y^{-1/3}y'=0\Rightarrow y'=-\left(\dfrac{y}{x}\right)^{1/3}$.
   4. Since $x=y$ here, $y'=-1$. …

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2. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$p_1$ and $p_2$ are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve $x^{2/3}+y^{2/3}=a^{2/3}$ respectively. If $k_1p_1^2+k_2p_2^2=a^2$, then $k_1+k_2=$ (A) $7$ (B) $6$ (C) $5$ (D) $4$

   ›Reveal solutionSolution

   Working out the tangent/normal distance formulas for the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ gives the identity $4p_1^2+p_2^2=a^2$, so $k_1+k_2=5$ — (C).

   Concept and Intuition

   Parametrizing the astroid trigonometrically turns the tangent and normal equations into simple linear equations in $x,y$ whose distance from the origin reduces to clean trig expressions in $\theta$. The key insight is that these expressions combine into a Pythagorean-type identity ($\sin^2+\cos^2=1$) once weighted correctly.

   Step-by-Step Solution

   1. Parametrize: $x=a\cos^3\theta,\ y=a\sin^3\theta$.
   2. Slope: $\dfrac{dy}{dx}=\dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\tan\theta$.
   3. Tangent line: $y-y_0=-\tan\theta(x-x_0)$. Multiplying by $\cos\theta$ and simplifying (using $x_0=a\cos^3\theta,y_0=a\sin^3\theta$) gives $\sin\theta\,x+\cos\theta\,y=a\sin\theta\cos\theta$. Distance from origin: $p_1=a|\sin\theta\cos\theta|=\dfrac a2|\sin2\theta|$.
   4. Normal line: slope $=\cot\theta$; similarly reduces to $\cos\theta\,x-\sin\theta\,y=a\cos2\theta$. Distance from origin: $p_2=a|\cos2\theta|$. …

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3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{2/3} + y^{2/3} = 4$ is parallel to the line $\sqrt{3}x + y = 1$, then $\alpha^2 + \beta^2 =$ (A) $10$ (B) $9$ (C) $28$ (D) $19$

   ›Reveal solutionSolution

   Matching the tangent slope of the astroid $x^{2/3}+y^{2/3}=4$ to the line's slope $-\sqrt3$ pins down the ratio $\beta/\alpha$; substituting back into the curve equation gives $\alpha^2+\beta^2=28$.

   Concept and Intuition

   For an implicit curve, differentiate both sides with respect to $x$ treating $y$ as a function of $x$, then solve for $dy/dx$. Setting that slope equal to a given line's slope (same coefficient structure $y=mx+c$) lets you find a relation between the point's coordinates.

   Step-by-Step Solution

   1. Differentiate $x^{2/3}+y^{2/3}=4$: $\tfrac23x^{-1/3}+\tfrac23y^{-1/3}\tfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\left(\dfrac{y}{x}\right)^{1/3}$.
   2. The line $\sqrt3x+y=1$ has slope $-\sqrt3$.
   3. At $(\alpha,\beta)$: $-\left(\dfrac{\beta}{\alpha}\right)^{1/3}=-\sqrt3 \Rightarrow \left(\dfrac{\beta}{\alpha}\right)^{1/3}=\sqrt3 \Rightarrow \dfrac{\beta}{\alpha}=3^{3/2}=3\sqrt3$.
   4. So $\beta=3\sqrt3\,\alpha$, hence $\beta^{2/3}=(3\sqrt3)^{2/3}\alpha^{2/3}=3\,\alpha^{2/3}$ (since $(3\sqrt3)^{2/3}=(3^{3/2})^{2/3}=3^1=3$). …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\dfrac{y}{x}\cos^4\alpha + \dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha \cdot \cos^2\alpha$, then $\dfrac{dy}{dx} =$ (A) $\sin^3\alpha \cos\alpha$ (B) $\sin^2\alpha \cos^2\alpha$ (C) $\dfrac{\sin^2\alpha}{\cos^2\alpha}$ (D) $\sin\alpha \cos^3\alpha$

   ›Reveal solutionSolution

   The given relation is secretly a perfect square in disguise; it forces $y=x\tan^2\alpha$, so $dy/dx=\tan^2\alpha$.

   Concept and Intuition

   Rather than differentiating implicitly right away, it pays to recognise the algebraic structure first. Multiplying by $xy$ converts the equation into a quadratic in $x$ and $y$ that factors as a perfect square, revealing $y/x$ is actually a constant (independent of $x$), which makes the derivative trivial.

   Step-by-Step Solution

   1. Start from $\dfrac{y}{x}\cos^4\alpha+\dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha$.
   2. Multiply both sides by $xy$: $y^2\cos^4\alpha + x^2\sin^4\alpha = 2xy\sin^2\alpha\cos^2\alpha$.
   3. Rearrange: $y^2\cos^4\alpha - 2xy\sin^2\alpha\cos^2\alpha + x^2\sin^4\alpha = 0$.
   4. This is $(y\cos^2\alpha - x\sin^2\alpha)^2 = 0$, so $y\cos^2\alpha = x\sin^2\alpha$, i.e. $y = x\tan^2\alpha$. …

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5. 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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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 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\sqrt{x-xy}+\sqrt{y-xy}=1$, then $\dfrac{dy}{dx}=$ (A) $-\sqrt{\dfrac{y-y^2}{x-x^2}}$ (B) $-\sqrt{\dfrac{1-y^2}{1-x^2}}$ (C) $-\sqrt{\dfrac{1-y}{1-x}}$ (D) $-\sqrt{\dfrac{x-y}{x+y}}$

   ›Reveal solutionSolution

   Squaring the constraint reveals that it forces $x+y=1$ identically, so $\dfrac{dy}{dx}=-1$ throughout; matching this against the options singles out $-\sqrt{\dfrac{y-y^2}{x-x^2}}$, since it equals $-1$ for every point satisfying $y=1-x$.

   Concept and Intuition

   Rather than blindly grinding through implicit differentiation of two square roots, it pays to first understand the curve itself. Squaring $\sqrt{x-xy}+\sqrt{y-xy}=1$ carefully (using $s=x+y,\,p=xy$) collapses to a perfect square equalling zero, revealing that the relation is nothing but the straight line $x+y=1$. Once we know that, $\dfrac{dy}{dx}=-1$ is immediate, and we just need to find which option reduces to $-1$ on this line.

   Step-by-Step Solution

   1. Square the given equation:

   $x(1-y)+y(1-x)+2\sqrt{xy(1-x)(1-y)} = 1$

   $x+y-2xy+2\sqrt{xy(1-x)(1-y)} = 1$

   2. Let $s=x+y,\ p=xy$. Then:

   $2\sqrt{p(1-x)(1-y)} = 1-s+2p$

   Note $(1-x)(1-y)=1-s+p$. Squaring again:

   $4p(1-s+p) = (1-s+2p)^2$

   Let $q=1-s$. Expanding both sides: LHS $=4pq+4p^2$; RHS $=q^2+4pq+4p^2$. So $0=q^2$, i.e. $q=0$, i.e. $s=1$.

   3. Hence $x+y=1$ is forced — the given relation is the line $y=1-x$ (restricted to the domain where the square roots are real, $0\le x,y\le1$).

   4. Differentiating $y=1-x$ directly: $\dfrac{dy}{dx}=-1$.

   5. Check which option gives $-1$ identically along $y=1-x$:

   • (A): $y-y^2=y(1-y)$. Substituting $y=1-x$: $y(1-y)=(1-x)\cdot x=x-x^2$. So the ratio is exactly $1$, and $-\sqrt1=-1$ for every $x$. ✓ …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $x^2 + y^2 = t - \dfrac{1}{t}$ and $x^4 + y^4 = t^2 + \dfrac{1}{t^2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y}{x}$ (B) $\dfrac{y^2}{x^2}$ (C) $\sqrt{\dfrac{y}{x}}$ (D) $-\dfrac{y}{x}$

   ›Reveal solutionSolution

   Eliminating the parameter $t$ between the two given equations produces the direct relation $x^2y^2=-1$ between $x$ and $y$, whose implicit derivative is $-y/x$.

   Concept and Intuition

   When $x$ and $y$ are both linked to a parameter $t$ through two equations, differentiating each with respect to $t$ separately (and dividing) works, but it is often faster — and here it is exact — to first eliminate $t$ algebraically to get a direct $x$–$y$ relation, then differentiate that implicitly in the ordinary way.

   Step-by-Step Solution

   1. Square the first equation: $(x^2+y^2)^2=\left(t-\dfrac1t\right)^2=t^2-2+\dfrac1{t^2}$, i.e. $x^4+2x^2y^2+y^4=t^2+\dfrac1{t^2}-2$.
   2. The second equation says $x^4+y^4=t^2+\dfrac1{t^2}$. Substitute this in: $\left(t^2+\dfrac1{t^2}\right)+2x^2y^2=t^2+\dfrac1{t^2}-2$.
   3. This forces $2x^2y^2=-2\Rightarrow x^2y^2=-1$ — a $t$-free relation directly linking $x$ and $y$. …

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9. 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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10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If $(a + \sqrt{2}\,b\cos x)(a - \sqrt{2}\,b\cos y) = a^2 - b^2$ where $a > b > 0$, then at $\left(\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$, $\dfrac{dy}{dx} =$ (A) $\dfrac{a + b}{a - b}$ (B) $\dfrac{a - b}{a + b}$ (C) $\dfrac{a - 2b}{a + 2b}$ (D) $\dfrac{2a + b}{2a - b}$

    ›Reveal solutionSolution

    Expand the product, simplify by dividing by the common factor $b$, then implicitly differentiate and evaluate at $x=y=\pi/4$. Answer: $\dfrac{a-b}{a+b}$.

    Concept and Intuition

    The given relation looks intimidating as a product, but expanding it cancels the $a^2$ on both sides (since the RHS is $a^2-b^2$) and leaves a much simpler equation relating $\cos x,\cos y$, and $\cos x\cos y$. From there it's routine implicit differentiation; the special evaluation point $x=y=\pi/4$ is chosen because $\sin$ and $\cos$ coincide there, which cancels neatly.

    Step-by-Step Solution

    1. Expand: $a^2 - a\sqrt2 b\cos y + a\sqrt2 b\cos x - 2b^2\cos x\cos y = a^2-b^2$.
    2. Cancel $a^2$ from both sides: $\sqrt2 ab(\cos x-\cos y) - 2b^2\cos x\cos y = -b^2$.
    3. Divide through by $b$ (nonzero): $\sqrt2 a(\cos x-\cos y) - 2b\cos x\cos y + b = 0$.
    4. Differentiate implicitly w.r.t. $x$ (treat $y=y(x)$):

    $\sqrt2 a\left(-\sin x + \sin y\cdot y'\right) - 2b\left(-\sin x\cos y - \cos x\sin y\cdot y'\right) = 0.$

    5. Group $y'$ terms: $y'\left(\sqrt2 a\sin y + 2b\cos x\sin y\right) = \sqrt2 a\sin x - 2b\sin x\cos y$.
    6. So $y' = \dfrac{\sin x(\sqrt2 a - 2b\cos y)}{\sin y(\sqrt2 a + 2b\cos x)}$.
    7. At $x=y=\dfrac\pi4$: $\sin x=\sin y=\cos x=\cos y=\dfrac1{\sqrt2}$. Substitute: …

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11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.If the relation $p(\text{subnormal length}) = q(\text{subtangent length})^2$ holds true for the curve $by^2=(x+a)^3$, then the value of $\dfrac{p}{q}=$ (A) $\dfrac{8}{27}$ (B) $\dfrac{8b}{27}$ (C) $\dfrac{8}{27b}$ (D) $\dfrac{27}{8b}$

    ›Reveal solutionSolution

    Computing the subnormal and subtangent for $by^2=(x+a)^3$ and matching the given relation yields $p/q = 8b/27$.

    Concept and Intuition

    For a curve $y=f(x)$ at a point, the subnormal length is $|y\,y'|$ and the subtangent length is $|y/y'|$ — both standard quantities from the geometry of tangent/normal lines meeting the x-axis. The problem gives a relation between them scaled by unknown constants $p,q$; computing both quantities explicitly and comparing powers of $(x+a)$ lets the $(x+a)^2$ factor cancel, leaving a clean ratio $p/q$ in terms of $b$ only.

    Step-by-Step Solution

    1. Curve: $by^2=(x+a)^3$. Differentiate implicitly: $2by\,y' = 3(x+a)^2 \Rightarrow y' = \dfrac{3(x+a)^2}{2by}$.
    2. Subnormal $= y\cdot y' = y\cdot\dfrac{3(x+a)^2}{2by} = \dfrac{3(x+a)^2}{2b}$.
    3. Subtangent $= \dfrac{y}{y'} = y\cdot\dfrac{2by}{3(x+a)^2} = \dfrac{2by^2}{3(x+a)^2}$. Using $y^2 = \dfrac{(x+a)^3}{b}$ (from the curve equation): Subtangent $= \dfrac{2b}{3(x+a)^2}\cdot\dfrac{(x+a)^3}{b} = \dfrac{2(x+a)}{3}$.
    4. So (Subtangent)$^2 = \dfrac{4(x+a)^2}{9}$. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.If the locus of the points on the curve $x^3y^2+\dfrac{x^2}{y}=5$ at which the tangent is parallel to X-axis is $f(x,y)=0$, then the point that lies on this curve $f(x,y)=0$ is (A) $(2,\sqrt[3]{3})$ (B) $(\sqrt[3]{2},3)$ (C) $\left(-2,\dfrac{1}{\sqrt[3]{3}}\right)$ (D) $\left(-\sqrt[3]{2},\dfrac{1}{\sqrt[3]{3}}\right)$

    ›Reveal solutionSolution

    Setting $dy/dx=0$ in the implicit differentiation of the curve gives the locus condition $3xy^3+2=0$; testing the four points, only $\left(-2,3^{-1/3}\right)$ satisfies it.

    Concept and Intuition

    "Tangent parallel to the X-axis" means $dy/dx=0$ at that point. Differentiating the curve implicitly and substituting $y'=0$ eliminates the derivative, leaving a plain algebraic relation between $x$ and $y$ — this relation is the locus equation $f(x,y)=0$, and we just need to check which candidate point satisfies it.

    Step-by-Step Solution

    1. Curve: $x^3y^2+x^2y^{-1}=5$.
    2. Differentiate w.r.t. $x$: $3x^2y^2+x^3\cdot2yy'+2xy^{-1}+x^2(-y^{-2})y'=0$, i.e. $3x^2y^2+2x^3yy'+\dfrac{2x}{y}-\dfrac{x^2y'}{y^2}=0$.
    3. Set $y'=0$ (horizontal tangent): $3x^2y^2+\dfrac{2x}{y}=0$.
    4. Multiply through by $y$: $3x^2y^3+2x=0 \Rightarrow x(3xy^3+2)=0$. Since $x=0$ doesn't satisfy the original curve, $x\ne0$, so $3xy^3+2=0 \Rightarrow xy^3=-\dfrac23$.
    5. Test each option against $xy^3=-\dfrac23$:
       • (A) $(2,3^{1/3})$: $xy^3=2\cdot3=6$ ✗
       • (B) $(2^{1/3},3)$: $xy^3=2^{1/3}\cdot27$ ✗ …

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