The slope of normal to the parabola y = 4 x 2 − 2 drawn through the point ( 10 , − 1 )

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The slope of normal to the parabola $y = \frac{x ^{2}}{4} - 2$ drawn through the point $(10, - 1)$

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Solution:

$x^{2} = 4 (y + 2)$ is the given parabola

Any normal is $x = m (y + 2) - 2 m - m^{3}$ . If $(10, - 1)$ lies on this line then $10 = + m - 2 m - m^{3} \Rightarrow m^{3} + m + 10 = 0 \Rightarrow m = - 2$

Slope of normal $= 1/ m$ .

Stream:JEESubject:MathematicsTopic:ParabolaSubtopic:Normal to Parabola

⏱ 2mℹ️ Source: QB

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