If $p$ is a prime and $a,b$ are integers both coprime to $p$, then prove that $ax^2+by^2\equiv$1(mod $p$) has $p-(\frac{-ab}{p})$ solutions where $(\frac{a}{p})$ denotes the Legendre's symbol.
My Attempt
$y^2\equiv b^{-1}(1-ax^2)$ (mod $p$).Now $(\frac{b^{-1}}{p})=(\frac{b}{p})$.So the number of solutions of $ax^2+by^2\equiv 1$(mod $p$) is the same as the number of solutions of $(\frac{b}{p})(\frac{1-ax^2}{p})=1$ as $x$ ranges from $0$ to $p-1$.
I am unable to proceed much further.
Following your attempt, it suffices to compute
$$\sum_{x=0}^{p-1}\left(1+\left(\frac bp\right)\left(\frac{1-ax^2}p\right)\right)=p+\left(\frac{-ab}p\right)\sum_{x=0}^{p-1}\left(\frac{x^2-a^{-1}}p\right).$$
Moreover,
$$\begin{aligned}\sum_{x=0}^{p-1}\left(\frac{x^2-a^{-1}}p\right)&=\sum_{x=0}^{p-1} \left(1+\left( \frac {x}{p}\right)\right)\left(\frac {x-a^{-1}}{p}\right)\\ &=0+\sum_{x=0}^{p-1}\left(\frac {x^2-a^{-1}x}{p}\right)\\ &=\sum_{x=1}^{p-1}\left(\frac {1-(ax)^{-1}}{p}\right)\\ &=\sum_{x=1}^{p-1}\left(\frac {1+x}{p}\right)\\ &=-1+\sum_{x=0}^{p-1}\left(\frac {1+x}{p}\right)=-1 \end{aligned}$$
(see Sum of Legendre symbol for the proof of the first equation above).