Question

Let the function $f(x)=2 x^2-\log _e x, x>0$, be decreasing in $(0, a )$ and increasing in $(a, 4)$. A tangent to the parabola $y^2=4 a x$ at a point $P$ on it passes through the point $(8 a, 8 a-1)$ 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-

Answer 1

$ f^{\prime}(x)=4 x-\frac{1}{x} $
$ a=\frac{1}{2}$
Let $P \left( x _1, y _1\right)$ be any point on $y ^2=4 ax$
$\frac{1}{y_1}=\frac{3-y_1}{4-x_1} \Rightarrow y_1^2-6 y_1+8=0$
$ y_1=2,4 $
$ \Rightarrow P(8,4) $ as $P(2,2) $ rejected
Equation of normal at $ P$ .
$y-4=-4(x-8) $
$ \frac{x}{9}+\frac{y}{36}=1 $
$ \alpha=9, \beta=36 $
$\alpha+\beta=45$