If the line $y=4 x-5$ touches the curve $y^{2}=a x^{3}+b$ at the point $(2,3)$, then
If the line $y=4 x-5$ touches the curve $y^{2}=a x^{3}+b$ at the point $(2,3)$, then
$a=-2, b=-7$
$a=-2, b=7$
$a=2 \quad, b=-7$
$a=2, b=7$
Solution
Line $y=4 x-5$ has slope 4
We have $y^{2}=a x^{3}+b \Rightarrow 2 y \frac{d y}{d x}=3 a x^{2}$
$\begin{array}{l}
\therefore \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{3 \mathrm{ax}^{2}}{2 \mathrm{y}} \\
\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,3)}=\frac{3 \mathrm{a}(2)^{2}}{2(3)}=2 \mathrm{a}
\end{array}$
As per given condition, $2 \mathrm{a}=4 \Rightarrow \mathrm{a}=2$
Now $y^{2}=a x^{3}+b$ at point $(2,3)$ becomes
$9=8(2)+b \Rightarrow b=-7$
This problem can be alternatively solved as follows:
Put $y=4 x-5$ in $y^{2}=a x^{3}+b$
$\therefore(4 x-5)^{2}=a x^{3}+b$
When $x=2$, we get
$9=8 a+b \ldots(1)$
Now we will go by options
Put $\mathrm{a}=2, \mathrm{~b}=-7$ in equation $(1)$ we get
$8(a)-7=16-7=9$