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  • The normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends an angle x at the focus, where x =  <div class='qna-option

The normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends an angle x at the focus, where x =

 

Option: 1

\mathrm{60^{\circ}}


Option: 2

\mathrm{75^{\circ}}


Option: 3

90^{\circ}


Option: 4

\mathrm{120^{\circ}}


Answers (1)

Let the equation of the parabola is \mathrm{y^2=4 a x}. Let PQ be a normal chord of the parabola at \mathrm{P\left(a t^2, 2 a t\right)}. Since the ordinate and abscissa of P are equal, \mathrm{a t^2=2 a t \Rightarrow t=2}

We have to prove that \mathrm{a t\angle P S Q=90^{\circ}}, where S is the focus (a, 0)

Now since t = 2, equation of the normal \mathrm{P Q \text { is } v+t x=2 a t+a t^3}

\mathrm{\text { or } y+2 x=4 a+8 a=12 a}

\mathrm{\text { If the coordinates of } Q \text { are }\left(a t_1^2, 2 a t_1\right) \text {, then } t_1=-t-\frac{2}{t}=-3}

Since S(a, 0) is the focus, gradients of SP and SQ are 

\mathrm{\frac{2 a t-0}{a t^2-a}=\frac{2 t}{t^2-1} \text { and } \frac{2 a t_1-0}{a t_1^2-a}=\frac{2 t_1}{t_1^2-1}}

\mathrm{\text { where } t=2, t_1=-3 \text {. Hence their product is }}

\mathrm{\frac{2 t}{t^2-1} \times \frac{2 t_1}{t_1^2-1}=\frac{4}{2^2-1} \times \frac{-6}{(-3)^2-1}=\frac{-4 \times 6}{3 \times 8}=-1}

\mathrm{\Rightarrow \quad S P \text { and } S Q \text { are perpendicular to each other, i.e., } \angle P S Q=90^{\circ} \text {. }}

 

 

Posted by

Sumit Saini

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