Solveeit Logo

Question

Question: The tangent to the curve 2y<sup>3</sup> = ax<sup>2</sup> + x<sup>3</sup> drawn at the point (a, a) ...

The tangent to the curve 2y3 = ax2 + x3 drawn at the point

(a, a) cuts l, m intercepts on coordinate axes where

l2 + m2 = 16. Then |a| is equal to –

A

31

B

30

C

28

D

16

Answer

30

Explanation

Solution

(dydx)(a,a)\left( \frac{dy}{dx} \right)_{(a,a)} = 2ax+3x26y2\frac{2ax + 3x^{2}}{6y^{2}} = 5a26a2\frac{5a^{2}}{6a^{2}} = 56\frac{5}{6}

equation of tangent ® y – a = 56\frac{5}{6} (x – a)

Ž x(a5)\frac{x}{\left( –\frac{a}{5} \right)} + y(a6)\frac{y}{\left( \frac{a}{6} \right)} = 1

given (a5)2\left( –\frac{a}{5} \right)^{2} + (a6)2\left( \frac{a}{6} \right)^{2}= 1 Ž a2 = (25) (36)

|a| = 30