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Maths-

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Easy

Question

The equation of the normal to the curve $y equals e to the power of negative 2 vertical line x vertical line end exponent$ at the point where the curve cuts the line $x equals 1 divided by 2$ , is

$2 e (e x + 2 y) = e^{2} - 4$
$2 e (e x - 2 y) = e^{2} - 4$
$2 e (e y - 2 x) = e^{2} - 4$
None of these

The correct answer is: $2 e open parentheses e x minus 2 y close parentheses equals e to the power of 2 end exponent minus 4$

We have,
$y equals e to the power of negative 2 vertical line x vertical line end exponent equals open curly brackets table row cell e to the power of 2 x end exponent comma blank x less than 0 end cell row cell e to the power of negative 2 x end exponent comma blank x greater or equal than 0 end cell end table close$

The line $x equals 1 divided by 2$ cuts this curve at the point $P open parentheses 1 divided by 2 comma blank e to the power of negative 1 end exponent close parentheses$

Also, $fraction numerator d y over denominator d x end fraction equals open curly brackets table row cell 2 e to the power of 2 x end exponent comma blank x less than 0 end cell row cell negative 2 e to the power of negative 2 x end exponent comma blank x greater than 0 end cell end table close$

$therefore open parentheses fraction numerator d y over denominator d x end fraction close parentheses subscript x equals 1 divided by 2 end subscript equals negative fraction numerator 2 over denominator e end fraction$

The equation of the normal at $P$ is

$y minus fraction numerator 1 over denominator e end fraction equals fraction numerator e over denominator 2 end fraction open parentheses x minus fraction numerator 1 over denominator 2 end fraction close parentheses$

$rightwards double arrow 4 open parentheses e y minus 1 close parentheses equals e to the power of 2 end exponent open parentheses 2 x minus 1 close parentheses rightwards double arrow 2 e open parentheses e x minus 2 y close parentheses equals e to the power of 2 end exponent minus 4$

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