If the tangent at any point on the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ intersects the co-ordinate axes in A and B, then show that the length AB is a constant.

If the tangent at any point on the curve x(2)/(3) + y(2)/(3) = a(2)/(3) intersects the co-ordinate axes in A a

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x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2024Subjective· 7mImportance★★★★★

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Find the tangent's slope by implicit differentiation, get its $x$ - and $y$ -intercepts $A,B$ in terms of the point of tangency, then show $OA^2+OB^2$ collapses to a constant using the curve's own equation.

Curve: $x^{2/3}+y^{2/3}=a^{2/3}$ . Differentiating implicitly:

$\frac23 x^{-1/3} + \frac23 y^{-1/3}\,y' = 0 \implies y' = -\left(\frac{y}{x}\right)^{1/3}$

At a point $(x_1,y_1)$ on the curve, the tangent is:

$y-y_1 = -\left(\frac{y_1}{x_1}\right)^{1/3}(x-x_1)$

$x$ -intercept $A$ (set $y=0$ ): solving gives $x = x_1^{1/3}\left(x_1^{2/3}+y_1^{2/3}\right) = x_1^{1/3}\,a^{2/3}$ , so $OA = a^{2/3}x_1^{1/3}$ .

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set SEM31 markMCQ
$y = x$
(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

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

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CBSE 2026Set SEM31 markMCQ
$lx - my + n = 0$
(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:

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

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CBSE 2024Set EX1 markMCQ
$y = x + 1$
(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:

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

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CBSE 2024Set ANNUAL1 mark
$y^2 - x^2 + 2x - 1 = 0$
›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 $x$ :

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

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CBSE 2024Set ANNUAL1 mark
$2y = 3 - x^3$
›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 $x=1$ : …

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

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CBSE 2023Set ANNUAL1 markMCQ
$y^2=x$
(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.

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 $x=0$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$f(x)=\sqrt{8-2x}$
(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 $x=-4$ .

$f(x)=\sqrt{8-2x}=(8-2x)^{1/2}$ . By the chain rule, $f'(x)=\dfrac12(8-2x)^{-1/2}\cdot(-2)=\dfrac{-1}{\sqrt{8-2x}}$ .

Set $f'(x)=-0.25=-\dfrac14$ : $\dfrac{-1}{\sqrt{8-2x}}=-\dfrac14\Rightarrow\sqrt{8-2x}=4$ . …
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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: $\dfrac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$ .

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CBSE 2022Set ANNUAL1 markMCQ
$y = \ln x\ (x>0)$
(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.

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

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CBSE 2022Set ANNUAL1 mark
$x =$
›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.

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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.

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

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←Previous chapterCh 11Continuity and Differentiability

Q.If the tangent at any point on the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ intersects the co-ordinate axes in A and B, then show that the length AB is a constant.

Concept understanding — Tangents and Normals (Lengths)

Tangents and Normals to a Curve

Q.If the tangent at any point on the curve x23+y23=a23x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}x32​+y32​=a32​ intersects the co-ordinate axes in A and B, then show that the length AB is a constant.

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Q.If the tangent at any point on the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ intersects the co-ordinate axes in A and B, then show that the length AB is a constant.

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