Question

There is a point $(p,q)$ on the graph of $f(x)=x^2$ and a point $(r,s)$ on the graph of $g(x)=\frac{-8}{x}$, where $p>$ 0 and $r>0$. If the line through $(p,q)$ and $(r,s)$ is also tangent to both the curves at these points, respectively, then the value of $p+r$ is

Answer 1

Answer: 5

$y=x^2$ and $y=-\frac{8}{x};q=p^2$ and $s=\frac{8}{r}...(i)$
Equation $\frac{dy}{dx}$ at $A$ and $B$, we get
$2p=\frac{8}{r^2}...(ii)$
$\Rightarrow p{r}^2=4$

Now $m_{AB}=\frac{q-s}{q-r}$
$\Rightarrow 2p=\frac{p^2+\frac{8}{r}}{p-r}$
$\Rightarrow p^2=2pr+\frac{8}{r}$
$\Rightarrow p^2=\frac{16}{r}$
$\Rightarrow \frac{16}{r^4}=\frac{16}{r}$
$\Rightarrow r=1(r\ne 0)$
$\Rightarrow p=4$
$\therefore r=1,p=1$
Hence $p+r=5$