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At a point (a// 8, a// 8) on the curve x...

At a point `(a// 8, a// 8)` on the curve `x^(1//3) + y^(1//3) = a^(1//3)` (a>0) tangent is drawn. If the portion of the tangent intercepted betweenthe axes be of length `sqrt2` then a=

A

1

B

2

C

4

D

8

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Find the slope of the tangent line We start with the curve given by the equation: \[ x^{1/3} + y^{1/3} = a^{1/3} \] To find the slope of the tangent line at the point \((\frac{a}{8}, \frac{a}{8})\), we differentiate the curve implicitly with respect to \(x\). Differentiating both sides: \[ \frac{1}{3} x^{-2/3} + \frac{1}{3} y^{-2/3} \frac{dy}{dx} = 0 \] Rearranging gives: \[ \frac{dy}{dx} = -\frac{y^{2/3}}{x^{2/3}} \] ### Step 2: Calculate the slope at the point \((\frac{a}{8}, \frac{a}{8})\) Substituting \(x = \frac{a}{8}\) and \(y = \frac{a}{8}\) into the slope formula: \[ \frac{dy}{dx} = -\frac{(\frac{a}{8})^{2/3}}{(\frac{a}{8})^{2/3}} = -1 \] ### Step 3: Write the equation of the tangent line Using the point-slope form of the equation of a line, we have: \[ y - \frac{a}{8} = -1 \left(x - \frac{a}{8}\right) \] Simplifying this gives: \[ y - \frac{a}{8} = -x + \frac{a}{8} \] Rearranging, we find: \[ y + x = \frac{a}{4} \] ### Step 4: Find the intercepts with the axes To find the x-intercept, set \(y = 0\): \[ 0 + x = \frac{a}{4} \implies x = \frac{a}{4} \] To find the y-intercept, set \(x = 0\): \[ y + 0 = \frac{a}{4} \implies y = \frac{a}{4} \] ### Step 5: Calculate the length of the intercepted segment The length of the segment intercepted between the axes is given by the distance formula between the points \((\frac{a}{4}, 0)\) and \((0, \frac{a}{4})\): \[ \text{Length} = \sqrt{\left(\frac{a}{4} - 0\right)^2 + \left(0 - \frac{a}{4}\right)^2} = \sqrt{\left(\frac{a}{4}\right)^2 + \left(\frac{a}{4}\right)^2} \] This simplifies to: \[ \text{Length} = \sqrt{2 \left(\frac{a}{4}\right)^2} = \sqrt{2} \cdot \frac{a}{4} = \frac{a \sqrt{2}}{4} \] ### Step 6: Set the length equal to \(\sqrt{2}\) According to the problem, this length is equal to \(\sqrt{2}\): \[ \frac{a \sqrt{2}}{4} = \sqrt{2} \] ### Step 7: Solve for \(a\) Multiplying both sides by 4: \[ a \sqrt{2} = 4 \sqrt{2} \] Dividing both sides by \(\sqrt{2}\): \[ a = 4 \] ### Conclusion The value of \(a\) is: \[ \boxed{4} \]
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ML KHANNA-TANGENTS AND NORMALS-PROBLEM SET (2) (MULTIPLE CHOICE QUESTIONS)
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  3. Out of the four curves given below chciose the curve which intersects...

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  4. The length of the subnormal to the parabola y^(2)=4ax at any point is ...

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  7. The length of the subtangent to the curve x^2+xy+y^2=7 at (1,-3) is

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  8. The length of the normal at t on the curve x=a(t+sint), y=a(1-cos t), ...

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  9. The length of the normal to the curve x= a(t +sin t),y = a(1-cos t), "...

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  10. Sum of squares of intercepts made on co-ordinate axes hy the tangents ...

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  11. The portion of the tangent of the curve x^(2/3)+y^(2/3)=a^(2/3) ,which...

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  12. At a point (a// 8, a// 8) on the curve x^(1//3) + y^(1//3) = a^(1//3) ...

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  13. In the curve x =a [cost+ log tan (t // 2)], y =a sin t, the portion of...

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  14. The triangle formed by the tangent to the curve f(x)=x^2+bx-b the poin...

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  15. The length of the normal at theta on the curve x = a cos^3 theta,y=asi...

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  16. The length of the normal to the curve at (x, y) y=a((e^(x//a)+e^(-x//a...

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  17. The value of n for which the length of the subnormal of the curve xy^n...

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  18. If the tangent at P on the curve x^my^n =d^(m+n) meets the co-ordinate...

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  19. For the parabola y^2 = 4ax, the ratio of the subtangentto the abscissa...

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  20. The tangent at any point. on the curve x^4+y^4=a^4 cuts off intercepts...

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