Question

If $y=mx+5$ is a tangent to $x^{3}y^{3}=ax^{3}+by^{3}$ at point $\left(1,2\right),$ then the value of $a$ is equal to

• A. $\frac{9}{5}$
• B. $\frac{16}{5}$
• C. $\frac{9}{4}$
• D. $\frac{18}{7}$

Answer 1

Answer: (B)

$P\left(1,2\right)$ lies on the curve and line, hence
$L \rightarrow 2=m+5\Rightarrow m=-3$
$Curve \rightarrow 8=a+8b$
Also, $\frac{d y}{d x}$ at $P\left(1,2\right)$ for the curve should be $m=-3,$
i.e. $3x^{2}y^{3}+x^{3}.3y^{2}y^{'}=3ax^{2}+3by^{2}y^{'}$
$\Rightarrow 24+12\left(- 3\right)=3a+12b\left(- 3\right)$
$\Rightarrow -12=3a-36b$
$\Rightarrow a-12b=-4$ & $a+8b=8$
$-20b=-12$
$\Rightarrow b=\frac{3}{5}$ and $a=\frac{16}{5}$