Let the function $f(x)=2 x^{2}-\log _{\mathrm{e}} x, x>0$, be decreasing in $(0, \mathrm{a})$ and increasing in $(\mathrm{a}, 4)$. A tangent to the parabola $y^{2}=4 a x$ at a point $\mathrm{P}$ on it passes through the point $(8 \mathrm{a}, 8 \mathrm{a}-1)$ but does not pass through the point $\left(-\frac{1}{a}, 0\right)$. If the equation of the normal at $P$ is : $\frac{x}{\alpha}+\frac{y}{\beta}=1$, then $\alpha+\beta$ is equal to ________________.

<p>$\delta '(x) = {{4{x^2} - 1} \over x}$ so f(x) is decreasing in $\left( {0,{1 \over 2}} \right)$ and increasing in $\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}$</p> <p>Tangent at ${y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}$</p> <p>It is passing through $(4,3)$</p> <p>$3 = 4m + {1 \over {2m}} \Rightarrow m = {1 \over 2}$ or ${1 \over 4}$</p> <p>So tangent may be</p> <p>$y = {1 \over 2}x + 1$ or $y = {1 \over 4}x + 2$</p> <p>But $y = {1 \over 2}x + 1$ passes through $( - 2,0)$ so rejected.</p> <p>Equation of normal</p> <p>$y = - 4x - 2\left( {{1 \over 2}} \right)( - 4) - {1 \over 2}{( - 4)^3}$</p> <p>or $y = - 4x + 4 + 32$</p> <p>or ${x \over 9} + {y \over {36}} = 1$</p>