JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 20)
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 ________________.
Answer
45
Explanation
$$\delta '(x) = {{4{x^2} - 1} \over x}$$ so f(x) is decreasing in $$\left( {0,{1 \over 2}} \right)$$ and increasing in $$\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}$$
Tangent at $${y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}$$
It is passing through $$(4,3)$$
$$3 = 4m + {1 \over {2m}} \Rightarrow m = {1 \over 2}$$ or $${1 \over 4}$$
So tangent may be
$$y = {1 \over 2}x + 1$$ or $$y = {1 \over 4}x + 2$$
But $$y = {1 \over 2}x + 1$$ passes through $$( - 2,0)$$ so rejected.
Equation of normal
$$y = - 4x - 2\left( {{1 \over 2}} \right)( - 4) - {1 \over 2}{( - 4)^3}$$
or $$y = - 4x + 4 + 32$$
or $${x \over 9} + {y \over {36}} = 1$$
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