AP EAMCET · Maths · Application of Derivatives
If the tangent of the curve \(4 y^3=3 a x^2+x^3\) drawn at the point \((a, a)\) forms a triangle of area \(\frac{25}{24}\) sq.units with the coordinate axes then \(\mathrm{a}=\)
- A \(\pm 10\)
- B \(\pm 5\)
- C \(\pm 6\)
- D \(\pm 3\)
Answer & Solution
Correct Answer
(B) \(\pm 5\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = \frac{6ax+3x^2}{12y^2}\) \(m = \left.\frac{dy}{dx}\right|_{(a,a)} = \frac{6a^2+3a^2}{12a^2} = \frac{9a^2}{12a^2} = \frac{3}{4}\) \(y - a = \frac{3}{4}(x - a) \implies 3x - 4y + a = 0\) x-intercept: \(-\frac{a}{3}\), y-intercept: \(\frac{a}{4}\) Area…
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