For the family of parabola $y=a_{n}x^2+b_{n}x+1$ and $a_{n}>0$ for all $n\in I^{+}$ and length of latus rectum of the parabolas are in H.P., also distance between the points where these parabolas cut the $x$-axis is always $1$. And $b_{1}^2, b_{2}^2, b_{3}^2.... $ are in A.P. with common difference equal to $\alpha$ times the common difference of A.P. corresponding to latus rectum of H.P., then $\alpha$ is
My solution: Latus rectum of $y=ax^2+bx+c$ is given by $\dfrac{1}{a}$ so $\dfrac{1}{a_{n}}$ are in H.P. hence $a_{1}, a_{2}, a_{3},...$ are in A.P with common difference let say $d$.
Since difference between roots of $y=a_{n}x^2+b_{n}x+1$ is always one. Let say roots are $x$ and $y$ so $|x-y|=1$. Hence $\sqrt{(x+y)^2-4xy}=1$ and using sum of roots as $x+y=\dfrac{-b_n}{a_n}$ and prodcut of root $xy=\dfrac{1}{a_n}$. I obtained $b_{n}^2=a_{n}^2+4a_{n}$ which does not form a A.P.
But given answer is $5$ and authur provided hint as $b_{n}^2=5a_{n}$
Please Help me solve this problem.