You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equation of Normal at Given Point

If P Q is a n...

Question

If $P Q$ is a normal chord of the parabola $y^{2} = 4 a x$ at $P (a t^{2}, 2 a t) .$ Then the axis of the parabola divides $¯ ¯¯¯¯¯¯ ¯ P Q$ in the ratio

\frac{t^{2}}{t^{2} + 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{t^{2}}{t^{2} - 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 t^{2}}{t^{2} - 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{t^{2}}{2 (t^{2} - 2)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{t^{2}}{t^{2} + 2}$
Let $P Q$ be normal chord, $P (a t^{2}, 2 a t), Q (a t_{2}^{2}, 2 a t_{2}), t_{2} = - t - \frac{2}{t}$
$As X-axis divides normal chord in the ratio - y_{1} : y_{2}$
$= \frac{- 2 a t}{2 a t_{2}} = \frac{t^{2}}{t^{2} + 2}$

Suggest Corrections

Similar questions

The chord AB of the parabola $y^{2} = 4 a x$ cuts the axis of the parabola at C. If $A = (a t_{1}^{2}, 2 a t_{1}), B = (a t_{2}^{2}, 2 a t_{2})$ and AC : AB = 1:3 then

Q. If the chord joining the points

t_{1}

and

t_{2}

to the parabola

y^{2} = 4 a x

is normal to the parabola at

t_{1}

then prove that

t_{1} (t_{1} + t_{2}) = - 2

Q. If the normal at

t_{1}

on the parabola

y^{2} = 4 a x

meets it again at

t_{2}

then

t_{2} =

Q. If the normal to the parabola

y^{2} = 4 a x

at the point

(a t^{2}, 2 a t)

cuts the parabola again at

(a T^{2}, 2 a T)

, then

Q. Let

P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r)

be three points on a parabola

y^{2} = 4 a x .

P Q

is the focal chord and

P K, Q R

are parallel where the co-ordinates of

K

(2 a, 0),

then the value of

r

Join BYJU'S Learning Program

If PQ is a normal chord of the parabola y2=4ax at P(at2,2at). Then the axis of the parabola divides ¯¯¯¯¯¯¯¯PQ in the ratio

The correct option is A t2t2+2Let PQbe normal chord,P(at2,2at),Q(at22,2at2),t2=−t−2t As X-axis divides normal chord in the ratio −y1:y2 =−2at2at2=t2t2+2

If $P Q$ is a normal chord of the parabola $y^{2} = 4 a x$ at $P (a t^{2}, 2 a t) .$ Then the axis of the parabola divides $¯ ¯¯¯¯¯¯ ¯ P Q$ in the ratio

The correct option is A $\frac{t^{2}}{t^{2} + 2}$
Let $P Q$ be normal chord, $P (a t^{2}, 2 a t), Q (a t_{2}^{2}, 2 a t_{2}), t_{2} = - t - \frac{2}{t}$
$As X-axis divides normal chord in the ratio - y_{1} : y_{2}$
$= \frac{- 2 a t}{2 a t_{2}} = \frac{t^{2}}{t^{2} + 2}$