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

If a parabola, whose latus rectum is

4 c,

slides between two rectangular axes, prove that the locus of its focus is

x^{2} y^{2} = c^{2} (x^{2} + y^{2})

, and that the curve traced out by its vertex is

x^{\frac{2}{3}} y^{\frac{2}{3}} (x^{\frac{2}{3}} + y^{\frac{2}{3}}) = c^{2}

.

Accepted Answer

Let $S$ be the focus and $O$ be region.

$P & Q$ are points of contact of the parabola with axes.

Since triangle $O S P$ and $Q S O$ similar, $∠ S O P = ∠ S Q O$ .

Hence, if we describe a circle through $O, Q$ and $S$ , then by geometry, $O P$ is the tangent to it at $O$ .

Hence, $S$ lies on the circle passing through the origin $O$ , the point $Q (0, b)$ and touching the axis of $X$ at the $(0, 0)$ .

The equation of this circle is $x^{2} + 2 x y \cos ω + y^{2} = b y$ .

Similarly, since $∠ S O Q = ∠ S P O$ will lie on the circle through $O$ and $P$ and touch the axis of $y$ at the origin on the circle,

$x^{2} + 2 x y \cos ω + y^{2} = b y . . . (1)$

And since $∠ S O Q = ∠ S P Q$ will be on the circle through $O$ and $P$ and touch the axis of $y$ at the origin on the circle,

$x^{2} + y^{2} + 2 x y \cos ω = a x . . . (2)$

The intersection of $(1)$ and $(2)$ we have towards the point

$(\frac{a b^{2}}{a^{2} + 2 a b \cos ω + b^{2}}, \frac{a^{2} b}{a^{2} + 2 a b \cos ω + b^{2}})$

$ω = 90 ° \Rightarrow \cos ω = 0$

So, focus is given by $x^{2} + y^{2} = a x = b y$ .

Eliminating $a$ and $b$ from $(1) & (2)$ we get,

$c^{2} {(x^{2} + y^{2})}^{6} {(\frac{1}{x^{2}} + \frac{1}{y^{2}})}^{3} = \frac{{(x^{2} + y^{2})}^{8}}{x^{2} y^{2}}$

$\Rightarrow x^{2} y^{2} = c^{2} (x^{2} + y^{2}) . . . (3)$

Coordinates of $x = \frac{a b^{y}}{{(a^{2} + b^{2})}^{2}}, y = \frac{a^{y} b}{{(a^{2} + b^{2})}^{2}}$

$\frac{x}{y} = \frac{b^{3}}{a^{3}} \Rightarrow \frac{b}{x^{\frac{1}{3}}} = \frac{a}{y^{\frac{1}{3}}} = λ$ (Suppose)

As $λ y^{\frac{1}{3}}$ and $b = x^{\frac{1}{3}}$ ,

putting the values of $a$ and $b$ in $(3)$ , we get

$C^{2} = \frac{λ^{2} x^{\frac{4}{3}} y^{\frac{4}{3}}}{{(x^{\frac{2}{3}} + y^{\frac{2}{3}})}^{2}}$ .

And eliminating $λ$ , we get $C^{2} = \frac{{(\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}})}^{2} x^{\frac{4}{3}} y^{\frac{4}{3}}}{{(x^{\frac{2}{3}} + y^{\frac{2}{3}})}^{2}}$ .

$\Rightarrow C^{2} = x^{\frac{2}{3}} y^{\frac{2}{3}} (x^{\frac{2}{3}} + y^{\frac{2}{3}})$

S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 3: EXAMPLES XXXI

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 3: EXAMPLES XXXI

Questions from S L Loney Solutions for Chapter: The Parabola (Continued), Exercise 3: EXAMPLES XXXI with Hints & Solutions

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