Let the function f(x) = 2x2 – logex, 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 _______ .
Let the function f(x) = 2x2 – logex, 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 _______ .
Correct Answer: 45
Solution and Explanation
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
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Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
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m×n = -1
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Diagram Explaining Tangents and Normal:
