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Standard XII

Mathematics

Sufficient Condition for an Extrema

PN is an ordi...

Question

PN is an ordinate of the parabola ; a straight line is drawn parallel to the axis to bisect NP and meets the curve in Q ; prove that NQ meets the tangent at the vertex in a point T such that $A T = \frac{2}{3} N P,$ where $A$ is the vertex.

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Solution

The equation of parabola be $y^{2} = 4 a x$

let the point $P$ be $(a t^{2}, 2 a t)$

$P N$ is ordinate $\Rightarrow N (a t^{2}, 0)$

Equation of straight line bisecting $N P$ is

$y = a t$

substituting $y$ in equation of parabola

$a^{2} t^{2} = 4 a x \Rightarrow x = \frac{a t^{2}}{4}$

So the coordinates of $Q$ are $(\frac{a t^{2}}{4}, a t)$

Equation of $N Q$ is

$y - 0 = \frac{a t - 0}{\frac{a t^{2}}{4} - a t^{2}} (x - a t^{2}) y = - \frac{4}{3 t} (x - a t^{2})$

Put $x = 0$

$y = - \frac{4}{3 t} (0 - a t^{2}) \Rightarrow y = \frac{4 a t}{3}$

$\Rightarrow A T = \frac{4 a t}{3} N P = 2 a t \frac{A T}{N P} = \frac{\frac{4 a t}{3}}{2 a t} = \frac{2}{3} A T = \frac{2}{3} N P$

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PN is an ordinate of the parabola ; a straight line is drawn parallel to the axis to bisect NP and meets the curve in Q ; prove that NQ meets the tangent at the vertex in a point T such that AT=23NP, where A is the vertex.

PN is an ordinate of the parabola ; a straight line is drawn parallel to the axis to bisect NP and meets the curve in Q ; prove that NQ meets the tangent at the vertex in a point T such that $A T = \frac{2}{3} N P,$ where $A$ is the vertex.