Question-29134

Question Number 29134 by Tinkutara last updated on 04/Feb/18

Commented by ajfour last updated on 04/Feb/18

Commented by ajfour last updated on 04/Feb/18

l e t P Q = r_{1}, P S = r_{2}, P R = r_{3}

\frac{1}{r_{2}} = \frac{r_{1} + r_{3}}{2 r_{1} r_{3}} \dots (i)

e q . o f v a r i a b l e l i n e t h r o u g h P :

y = β + r \sin θ, x = α + r \cos θ

Q ({a t}_{1}^{2}, 2 {a t}_{1}), R ({a t}_{2}^{2}, 2 {a t}_{2})

Q a n d R l i e o n l i n e a n d p a r a b o l a

\Rightarrow 2 a t = β + r \sin θ

{a t}^{2} = α + r \cos θ

{(β + r \sin θ)}^{2} = 4 a (α + r \cos θ)

\Rightarrow r^{2} \sin^{2} θ + (2 β \sin θ - 4 a \cos θ) r + β^{2} - 4 a α = 0

r_{1}, r_{3} a r e r o o t s o f t h i s e q .

\Rightarrow \frac{2 r_{1} r_{3}}{r_{1} + r_{3}} = \frac{2 (β^{2} - 4 a α)}{4 a \cos θ - 2 β \sin θ}

u s i n g t h i s i n (i)

\frac{1}{r_{2}} = \frac{4 a \cos θ - 2 β \sin θ}{2 (β^{2} - 4 a α)} \dots (i i)

o r β^{2} - 4 a α = 2 {a r}_{2} \cos θ - β r_{2} \sin θ

b u t S (h, k) l i e s o n l i n e, s o

h = α + r_{2} \cos θ, k = β + r_{2} \sin θ

u s i n g t h i s i n (i i)

\boldsymbol β^{2} - 4 \boldsymbol a α = 2 \boldsymbol a (\boldsymbol h - \boldsymbol α) - \boldsymbol β (\boldsymbol k - \boldsymbol β)

\Rightarrow 2 a (h + α) = k β

s o r e p l a c i n g (h, k) b y (x, y)

\boldsymbol y = (\frac{2 \boldsymbol a}{\boldsymbol β}) x + \frac{2 \boldsymbol a α}{\boldsymbol β} .

\boldsymbol s l o p e \boldsymbol i n d e p e n d e n t \boldsymbol o f \boldsymbol α .

Commented by Tinkutara last updated on 05/Feb/18

Wonderful Sir! �� Thanks! ��