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Directrix of Ellipse

Find the leng...

Question

Find the length of normal to $y^{2} = 4 a x$ at $(a t^{2}, 2 a t)$

A

\sqrt{(t^{2} - 1}

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B

a \sqrt{(t^{2} + 1}

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C

2 a \sqrt{(t^{2} - 1}

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D

2 a \sqrt{(t^{2} + 1}

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Solution

The correct option is B $2 a \sqrt{(t^{2} + 1}$
Equation of normal at $P (a t^{2}, 2 a t)$ is
$y = - t x + 2 a t + a t^{3}$
Put $y = 0$
$\Rightarrow 0 = - t x + 2 a t + a t^{3} \Rightarrow x = 2 a + a t^{2}$
So, it intersect the axis at $N (2 a + a t^{2}, 0)$ .
Lenght of normal $= P N$
$\Rightarrow P N = \sqrt{{(a t^{2} - 2 a + a t^{2})}^{2} + {(2 a t - 0)}^{2}} \Rightarrow P N = \sqrt{4 a^{2} + 4 a^{2} t^{2}} \Rightarrow P N = 2 a \sqrt{1 + t^{2}}$
So, option D is correct.

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Similar questions

A = (a t^{2}, 2 a t), B = (\frac{a}{t^{2}}, - \frac{2 a}{t}), S (a, 0)

\frac{1}{S A} + \frac{1}{S B} =

Q. Consider a parabola

y^{2} = 4 a x .

If the normal to the parabola at the point

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

cuts the parabola again at

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

Q. If the normal to the parabola

y^{2} = 4 a x

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

cuts the parabola again at

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

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

Q. Match the column
The parabola

y^{2} = 4 a x

A B

A (a t_{1}^{2}, 2 a t_{1})

B (a t_{2}^{2}, 2 a t_{2})

Column - I	Column - II
(A) $A B$ is a normal chord if	(P) $t_{2} = - t_{1} - \frac{1}{2}$
(B) $A B$ is a focal chord	(Q) $t_{2} = \frac{- 4}{t_{2}}$
(C) $A B$ subtends $90^{\circ}$ at point $(0, 0)$ if	(R) $t_{2} = \frac{- 1}{t_{1}}$ (S) $t_{2} = - t_{1} - \frac{2}{t_{1}}$

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