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Byju's Answer

Standard XII

Mathematics

Slope Form of Tangent

A line bisect...

Question

$A$ line bisecting the ordinate $P N$ of a point $P (a t^{2}, 2 a t), t > 0$ , on the parabola $y^{2} = 4 a x$ is drawn parallel to the axis to meet the curve at $Q$ . lf $N Q$ meets the tangent at the vertex at the point $T$ . then the coordinates of $T$ are

(0, \frac{4}{3} a t)

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(0, 2 a t)

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(\frac{1}{4} a t^{2}, a t)

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(0, a t)

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Solution

The correct option is A $(0, \frac{4}{3} a t)$

$a^{2} t^{2} = 4 a x$
$x = \frac{a t^{2}}{4}$
$(y - 0) = ⎛ ⎜ ⎝ \frac{a t - 0}{\frac{a t^{2}}{4} - a t^{2}} ⎞ ⎟ ⎠ (x - a t^{2})$
put $x = 0$
$y = \frac{a t .4}{3 a t^{2}} \times a t^{2}$

$y = \frac{4 a t}{3}$

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Similar questions

Q. A line bisecting the ordinate

P N

of a point

P (a t^{2}, 2 a t), t > 0

, on the parabola

y^{2} = 4 a x

is drawn parallel the axis to meet the curve at

Q

. if

N Q

meets the tangent at the vertex at the point

T

, then the coordinates of

T

are

Q. A line bisecting the ordinate

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of a point

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on the parabola

y^{2} = 4 a x

is drawn parallel to the axis to meet the curve at

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. If

N Q

meets the tangents at the vertex at the point

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, then the coordinates of

T

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is an ordinate of the parabola

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parallel to the axis of the parabola meeting the parabola at

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meets the tangent at the vertex

A,

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\frac{A T}{N P} =

Q. 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

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A line bisecting the ordinate PN of a point P(at2,2at),t>0, on the parabola y2=4ax is drawn parallel to the axis to meet the curve at Q. lf NQ meets the tangent at the vertex at the point T. then the coordinates of T are

The correct option is A (0,43at)a2t2=4axx=at24(y−0)=⎛⎜⎝at−0at24−at2⎞⎟⎠(x−at2)put x=0y=at.43at2×at2y=4at3

$A$ line bisecting the ordinate $P N$ of a point $P (a t^{2}, 2 a t), t > 0$ , on the parabola $y^{2} = 4 a x$ is drawn parallel to the axis to meet the curve at $Q$ . lf $N Q$ meets the tangent at the vertex at the point $T$ . then the coordinates of $T$ are

The correct option is A $(0, \frac{4}{3} a t)$

$a^{2} t^{2} = 4 a x$
$x = \frac{a t^{2}}{4}$
$(y - 0) = ⎛ ⎜ ⎝ \frac{a t - 0}{\frac{a t^{2}}{4} - a t^{2}} ⎞ ⎟ ⎠ (x - a t^{2})$
put $x = 0$
$y = \frac{a t .4}{3 a t^{2}} \times a t^{2}$

$y = \frac{4 a t}{3}$