Question

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

Answer: 45

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