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Maths-

General

Easy

Question

The normal y = mx – 2am – am² to the parabola y² = 4ax subtends a right angle at the origin, then-

m = 1
m = $\sqrt{2}$
m = 2
m = $\frac{1}{\sqrt{2}}$

The correct answer is: m = $square root of 2$

Standard equation of the normal at P(at², 2at) is y + xt = 2at + at3.
Hence t = – m. This meets the parabola again at Q whose parameter is – t – $fraction numerator 2 over denominator t end fraction$ or m + $fraction numerator 2 over denominator m end fraction$
$therefore$ coordinates of Q are a $open parentheses m plus fraction numerator 2 over denominator m end fraction close parentheses to the power of 2 end exponent$ , 2a $open parentheses m plus fraction numerator 2 over denominator m end fraction close parentheses$
Since $angle P O Q equals 90 to the power of ring operator end exponent rightwards double arrow open parentheses fraction numerator 2 over denominator m plus fraction numerator 2 over denominator m end fraction end fraction close parentheses open parentheses fraction numerator 2 over denominator negative m end fraction close parentheses equals negative 1$
$table row cell rightwards double arrow m to the power of 2 end exponent plus 2 equals 4 end cell row cell rightwards double arrow m equals plus-or-minus square root of 2 end cell end table$
m = $square root of 2$ is one of the choices.

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