JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 19)
If the point on the curve y2 = 6x, nearest to the point $$\left( {3,{3 \over 2}} \right)$$ is ($$\alpha$$, $$\beta$$), then 2($$\alpha$$ + $$\beta$$) is equal to _____________.
Answer
9
Explanation
Let, $$P \equiv \left( {{3 \over 2}{t^2},3t} \right)$$ is on the curve.
Normal at point P
$$tx + y = 3t + {3 \over 2}{t^3}$$
Passes through $$\left( {3,{3 \over 2}} \right)$$
$$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$$
$$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$$
$$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$$
$$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$$
Normal at point P
$$tx + y = 3t + {3 \over 2}{t^3}$$
Passes through $$\left( {3,{3 \over 2}} \right)$$
$$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$$
$$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$$
$$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$$
$$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$$
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