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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

125. The points on the curve 9y² = x³ , where the normal to the curve makes equal intercepts with the axes are

(A) $(4, \pm \frac{8}{3})$ (B) $4, \frac{- 8}{3}$ (C) $4, \pm \frac{3}{8}$ (D) $\pm 4, \frac{3}{8}$

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Vishal Baghel | Contributor-Level 10

4 months ago

The equation of the given curve is $9 y^{2} = x^{3}$

Differentiate with respect to x, we have:

$\begin{array}{l} 9 (2 y) \frac{d y}{d x} = 3 x^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{x^{2}}{6 y} \end{array}$

The slope of the normal to the given curve at point $(x_{1}, y_{1})$ is

$\frac{- 1}{{\frac{d y}{d x}]}_{(x_{1}, y_{1})}} = - \frac{6 y_{1}}{{x_{1}}^{2}}$

$∴$ The equation of the normal to the curve at $(x_{1}, y_{1})$ is

$\begin{array}{l} y - y_{1} = - \frac{6 y_{1}}{{x_{1}}^{2}} (x - x_{1}) \\ \Rightarrow {x_{1}}^{2} y - {x_{1}}^{2} y_{1} = - 6 x y_{1} + 6 x_{1} y_{1} \\ \Rightarrow 6 x y_{1} + {x_{1}}^{2} y = 6 x_{1} y_{1} + {x_{1}}^{2} y_{1} \\ \Rightarrow \frac{6 x y_{1}}{6 x_{1} y_{1} + {x_{1}}^{2} y_{1}} + \frac{{x_{1}}^{2} y}{6 x_{1} y_{1} + {x_{1}}^{2} y_{1}} = 1 \\ \Rightarrow \frac{x}{\frac{x_{1} (6 + x_{1})}{6}} + \frac{y}{\frac{y_{1} (6 + x_{1})}{x_{1}}} \end{array}$

It is given that the normal makes intercepts with the axes.

Therefore, we have:

$\begin{array}{l} ∴ \frac{x_{1} (6 + x_{1})}{6} = \frac{y_{1} (6 + x_{1})}{x_{1}} \\ \Rightarrow \frac{x_{1}}{6} = \frac{y_{1}}{x_{1}} \\ \Rightarrow {x_{1}}^{2} = 6 y_{1} . . . . . . . . . . (i) \end{array}$

Also, the point $(x_{1}, y_{1})$ lies on the curve, so we have

$9 {y_{1}}^{2} = {x_{1}}^{3} . . . . . . . . . . (i i)$

From (i) and (ii), we have:

$9 (\frac{{x_{1}}^{2}}{6}) = {x_{1}}^{3} \Rightarrow \frac{{x_{1}}^{4}}{4} = {x_{1}}^{3} \Rightarrow x_{1} = 4$

From (ii), we have:

$\begin{array}{l} 9 {y_{1}}^{2} = {(4)}^{3} = 64 \\ \Rightarrow {y_{1}}^{2} = \frac{64}{9} \\ \Rightarrow y_{1} = \pm \frac{8}{3} \end{array}$

Hence, the required points are $(4, \pm \frac{8}{3})$ .

Therefore, option (A) is correct.

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If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

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y (x) = ∫? (2t² - 15t + 10)dt
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2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

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From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

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So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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125. The points on the curve 9y² = x³ , where the normal to the curve makes equal intercepts with the axes are

(A) $(4, \pm \frac{8}{3})$ (B) $4, \frac{- 8}{3}$ (C) $4, \pm \frac{3}{8}$ (D) $\pm 4, \frac{3}{8}$

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125. The points on the curve 9y2 = x3 , where the normal to the curve makes equal intercepts with the axes are (A) (4,±83) (B) 4,−83 (C) 4,±38 (D) ±4,38

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125. The points on the curve 9y² = x³ , where the normal to the curve makes equal intercepts with the axes are

(A) $(4, \pm \frac{8}{3})$ (B) $4, \frac{- 8}{3}$ (C) $4, \pm \frac{3}{8}$ (D) $\pm 4, \frac{3}{8}$