Question

Let the tangent to $y=f\left(x\right)$ at $(a,f(a\left)\right)$ has $x-$intercept $(a-2)$ and $f\left(0\right)=2$. If $f\left(x\right)$ is of the form of $k{e}^{px}$, then the value of $\left(\frac{k}{p}\right)$ is

Answer 1

$y=k{e}^{px}$ passes through $(0,2)\Rightarrow k=2$$\therefore y=2e^{px}$
On substituting $x=a$, we have $y=2e^{ap}$
$\therefore (a,2e^{\text{ap}})$ is a point on curve
$y^{\prime }=2p{e}^{px}$
$y^{\prime }$ at $(a,2e^{ap})=2p{e}^{ap}$
$\therefore$ Equation of tangent is
$(y-2e^{ap})=2p{e}^{ap}(x-a)$
For $x-$intercept, put $y=0$
$\Rightarrow x=-\frac{1}{p}+a=a-2(\text{Given}x-\text{intercept})$
$\Rightarrow p=\frac{1}{2}$

$\therefore \frac{k}{p}=\frac{2}{\frac{1}{2}}=4$