Question

#Quality Question
@ Coordinate Geometry

The portion of tangent to the parabola y^2 = 4ax cut off between the directrix and the curve subtends an angle θ at the focus where θ = kπ
then k is . . .
​

Answer 1

Given:

• The portion of tangent to the parabola y² = 4ax cut off between the directrix and the curve
• subtends an angle θ at the focus where θ = kπ

Solution:

The equation of the tangent at

P ( at² ,2at ) to y² = 4ax is ty = x + at² -----(1)

It meets the direction x = -a

•°• ty = -a + at² ⟹ y = $\frac{a( {t}^{2} - 1)}{t}$

Thus , (1) meets the direction at

$Q( - a,\frac{a( {t}^{2} - 1) }{t} )$

Now slope of PS is $m _{1} = \frac{2at - 0}{at ^{2} - a } = \frac{2t}{ {t}^{2} - 1}$

And slope of QS is $m _{2} = \frac{a(t ^{2} - 1) \div (t - 0) }{a - a } = \frac { - (t ^{2} - 1 )}{ 2t}$

Since, $m _{1} \: m _{2}= - 1$, therefore PQ subtends a right angle at the focus.

Hence, $θ= \frac{\pi}{2}$

Answer 2

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ For the parabola y² = 4ax,

• The coordinates of focus is S(a, 0).

• The equation of directrix is x = - a.

↝ Let suppose that PQ be a tangent to the parabola y² = 4ax touches the parabola at P and intersects the directrix at Q.

↝ Let assume the coordinates of P in parametric form as P ( at² ,2at ), where t is parameter.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ The equation of the tangent in parametric form at P ( at² ,2at ) to y² = 4ax is

$\bf :\longmapsto\:ty = x + at^{2} - - - (1)$

↝ Since, Tangent PQ meets the dirextrix at Q.

↝ So, point of intersection of directrix and tangent provides the coordinates of Q.

↝ Equation of directrix is x = - a -----(2)

So, On substituting (2) in (1), we get

$\bf :\longmapsto\:ty = - a + at^{2}$

$\bf :\longmapsto\:ty = a(t^{2} - 1)$

$\bf\implies \:y = \dfrac{a( {t}^{2} - 1)}{t}$

Hence,

$\rm :\longmapsto\:Coordinates \: of \: Q = \bigg( - a, \: \dfrac{a( {t}^{2} - 1)}{t}\bigg)$

Now,

We have

$\rm :\longmapsto\:Coordinates \: of \: Q = \bigg( - a, \: \dfrac{a( {t}^{2} - 1)}{t}\bigg)$

$\rm :\longmapsto\:Coordinates \: of \: P= (a {t}^{2}, \: 2at)$

$\rm :\longmapsto\:Coordinates \: of \: S= (a, \: 0)$

Further we know that,

↝ The slope of line joining the points (a, b) and (c, d) is represented as m and given by

$\underbrace{ \boxed{ \bf \: m = \frac{d - b}{c - a} }}$

So, using this,

$\rm :\longmapsto\:Slope \: of \: PS, \: m_1 = \dfrac{2at - 0}{ {at}^{2} - a}$

$\rm :\longmapsto\:Slope \: of \: PS, \: m_1 = \dfrac{2at}{ {a(t}^{2} - 1)}$

$\rm :\longmapsto\:Slope \: of \: PS, \: m_1 = \dfrac{2t}{ {(t}^{2} - 1)} - - - (3)$

And Now,

$\rm :\longmapsto\:Slope \: of \: QS, \: m_2 = \dfrac{\dfrac{a( {t}^{2} - 1)}{t} - 0}{ - a - a}$

$\rm :\longmapsto\:Slope \: of \: QS, \: m_2 = \dfrac{\dfrac{a( {t}^{2} - 1)}{t}}{ -2a}$

$\rm :\longmapsto\:Slope \: of \: QS, \: m_2 = \dfrac{( {t}^{2} - 1)}{ - 2t} - - - (4)$

Now, it is given that

↝ Angle betwen PS and QS is kπ.

So, using angle between two lines QS and PS,

$\rm :\longmapsto\:tan \: k\pi = \bigg|\dfrac{m_1 - m_2}{1 + m_1m_2} \bigg|$

On Substituting the values from equation (3) and (4), we get

$\rm :\longmapsto\:tan \: k\pi = \bigg|\dfrac{\dfrac{2t}{( {t}^{2} - 1)} - \dfrac{( {t}^{2} - 1)}{2t}}{1 + ( - 1)} \bigg|$

$\rm :\longmapsto\:tan \: k\pi = \bigg|\dfrac{\dfrac{2t}{( {t}^{2} - 1)} - \dfrac{( {t}^{2} - 1)}{2t}}{1 - 1} \bigg|$

$\rm :\longmapsto\:tan \: k\pi = \bigg|\dfrac{\dfrac{2t}{( {t}^{2} - 1)} - \dfrac{( {t}^{2} - 1)}{2t}}{0} \bigg|$

$\rm :\longmapsto\:tan \: k\pi = \infty$

$\bf\implies \:k\pi = \dfrac{\pi}{2}$

$\bf\implies \:k = \dfrac{1}{2}$