Name: Pair of Tangent, Chord of Contact
Uploaded: 2019-07-19
Duration: 19 min 37 s
Description: This video contains all the details of the chord of the contact pair of tangent and cord bisected at a point. This concept is very useful for the students.

Question

Show that three circles can be drawn to touch a parabola and also to touch at the focus of a given straight line passing through the focus. Also prove that the tangents at the point of contact with the parabola form an equilateral triangle.

Accepted Answer

Let the equation of circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

We know that the equation to the circle, which passes through the focus $(a, 0)$ is

$a^{2} + 2 g a + c = 0 \Rightarrow c = - a^{2} - 2 g a . . . (i)$

Circle also touches the parabola $y^{2} = 4 a x$ at $(a t^{2}, 2 a t)$ . This implies slope of circle at this point will be $\frac{1}{t}$ .

Differentiating equation of circle we get,

$2 x + 2 y y^{'} + 2 g + 2 f y^{'} = 0$

$\Rightarrow 2 a t^{2} + 2 (2 a t) \frac{1}{t} + 2 g + 2 f \frac{1}{t} = 0$

$\Rightarrow a t^{3} + 2 a t + g t + f = 0 . . . (ii)$

Also point $(a t^{2}, 2 a t)$ satisfies equation of circle

$\Rightarrow a^{2} t^{4} + 4 a^{2} t^{2} + 2 g a t^{2} + 4 f a t + c = 0 . . . (iii)$

Solving $(i), (ii)$ and $(iii)$ , values of variables $g = - \frac{a}{2} (3 t^{2} + 1), f = - \frac{a}{2} (3 t - t^{3})$ and $c = 3 a^{2} t^{2}$ .

So, the equation of circle will be $x^{2} + y^{2} - a x (3 t^{2} + 1) - a y (3 t - t^{3}) + 3 a^{2} t^{2} = 0$ .

And equation of any line through $(a, 0)$ is

$y = \frac{\sin α}{\cos α} (x - a)$ $(∵ m = \tan α = \frac{\sin α}{\cos α})$ .

$\Rightarrow x \sin α - y \cos α = a \sin α$

So, the equation of perpendicular line will be

$x \cos α + y \sin α = a \cos α$ .

Co-ordinates of the center of the circle are $(a (\frac{3 t^{2} + 1}{2}), a (\frac{3 t - t^{3}}{2}))$ .

Since the line passes through the center, we have

$\frac{a (3 t^{2} + 1)}{2} \cos α + \frac{a}{2} (3 t - t^{3}) \sin α = a \cos α$

Or $(3 t^{2} + 1) \cos α + (3 t - t^{3}) \sin α = 2 \cos α$

Or $(3 t - t^{3}) \sin α = (1 - 3 t^{2}) \cos α$

Putting $t = \tan θ$ , we get

$\cot α = \frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ} = \tan 3 θ$

Or $\tan 3 θ = \tan (\frac{π}{2} - α)$

Solving, we have $3 θ = (\frac{π}{2} - α), \{π + (\frac{π}{2} - α)\}$ or $\{2 π + (\frac{π}{2} - α)\}$

If $θ_{1}, θ_{2}$ and $θ_{3}$ be three values of $θ$ , then

$θ_{1} = \frac{1}{3} (\frac{π}{2} - α), θ_{2} = \frac{π}{3} + \frac{1}{3} (\frac{π}{2} - α)$ and $θ_{3} = \frac{2 π}{3} + \frac{1}{3} (\frac{π}{2} - α)$

So, $θ_{2} - θ_{1} = \frac{π}{3} = θ_{3} - θ_{2}$ .

Since the three normals are inclined at $60 °$ , the angles between the tangents are also $60 ° .$

Therefore, the three normals form an equilateral triangle.

Three different values implies three circles can be drawn.

Hence, proved.

Show that three circles can be drawn to touch a parabola and also to touch at the focus of a given straight line passing through the focus. Also prove that the tangents at the point of contact with the parabola form an equilateral triangle.

Important Questions on The Parabola (Continued)

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