Question

Let the function f(x) = 2x² – log_ex, x> 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y² = 4ax at a point P on it passes through the point (8a, 8a –1) but does not pass through the point (-1/a, 0). If the equation of the normal at P is
\(\frac{x}{α}+\frac{y}{β}=1\)
 then α + β is equal to _______ .

Answer 1

Correct Answer: 45

The correct answer is 45
\(δ^′(x)=\frac{4x^2−1}{x}\)
so f(x) is decreasing in \((0,\frac{1}{2})\) and increasing in \((\frac{1}{2},∞)\)
\(⇒a=\frac{1}{2}\)
Tangent at \(y^2=2x\)
\(⇒y=mx+\frac{1}{2m}\)
It is passing through (4, 3)
\(3=4m+\frac{1}{2m}\)
\(⇒m=\frac{1}{2} or \frac{1}{4}\)
So tangent may be
\(y=\frac{1}{2}x+1 or\ y=\frac{1}{4}x+2\)
But \(y=\frac{1}{2}x+1\) passes through (–2, 0) so rejected.
Equation of Normal
\(y=−4x−2(\frac{1}{2})(−4)−\frac{1}{2}(−4)^3\)
\(y=−4x+4+32\)
\(\frac{x}{9}+\frac{y}{36}=1\)
 α + β = 9 + 36
= 45