Correct option is (A) 8
\((2x-3y)\) \((4x^2+6xy+9y^2)\)
\(\because\) \(x^3\) will get by multiplying 2x with \(4x^2.\)
\(\therefore\) Coefficient of \(x^3\) in \((2x-3y)\) \((4x^2+6xy+9y^2)\) is \(2\times4=8.\)
\(\frac{a^2+6}{a^3-8}\) x \(\frac{2a^2-3a-2}{2a^2+9a+4}\) ÷ \(\frac{3a^2-11a-4}{a^2+2a+4}\)
= \(\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\) x \(\frac{(2a+1)(a-2)}{(2a+1)(a+4)}\) x \(\frac{(a+2)^2}{(3a+1)(a-4)}\)
= \(\frac{(a+2)^2}{(a^2+2a+4)(3a+1)}\)
`24x^(2)-6xy+29y+6x-58y-151=0`
`2(x-3y+3)^(2)+2(3x+y-1)^(2)=180`
or `((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1`
or `((x-3y+3)/(sqrt(1+3^(2))sqrt(6)))^(2)+((3x+y-1)/(3sqrt(1+3^(2))))=1`
Thus, C is an ellipse whose lengths of axes are `6,2sqrt(6)`.
The minor and the major axes are `x-3y+3=0 and 3x+y-1=0`, respectively.
Their point...
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