Given parabola y 2 = 4 x {y^2} = 4x y 2 = 4 x
∴ \therefore ∴ a = 1
Here, P, S, Q points are collinear.
∴ \therefore ∴ Slope of PS = Slope of QS
⇒ 2 t 1 − 0 t 1 2 − 1 = 0 − 2 t 2 1 − t 2 2 \Rightarrow {{2{t_1} - 0} \over {t_1^2 - 1}} = {{0 - 2{t_2}} \over {1 - t_2^2}} ⇒ t 1 2 − 1 2 t 1 − 0 = 1 − t 2 2 0 − 2 t 2
⇒ 2 t 1 t 1 2 − 1 = 2 t 2 t 2 2 − 1 \Rightarrow {{2{t_1}} \over {t_1^2 - 1}} = {{2{t_2}} \over {t_2^2 - 1}} ⇒ t 1 2 − 1 2 t 1 = t 2 2 − 1 2 t 2
⇒ t 1 ( t 2 2 − 1 ) = t 2 ( t 1 2 − 1 ) \Rightarrow {t_1}(t_2^2 - 1) = {t_2}(t_1^2 - 1) ⇒ t 1 ( t 2 2 − 1 ) = t 2 ( t 1 2 − 1 )
⇒ t 2 2 t 1 − t 1 = t 1 2 t 2 − t 2 \Rightarrow t_2^2{t_1} - {t_1} = t_1^2{t_2} - {t_2} ⇒ t 2 2 t 1 − t 1 = t 1 2 t 2 − t 2
⇒ t 2 2 t 1 − t 1 2 t 2 − t 1 + t 2 = 0 \Rightarrow t_2^2{t_1} - t_1^2{t_2} - {t_1} + {t_2} = 0 ⇒ t 2 2 t 1 − t 1 2 t 2 − t 1 + t 2 = 0
⇒ t 1 t 2 ( t 2 − t 1 ) + ( t 2 − t 1 ) = 0 \Rightarrow {t_1}{t_2}({t_2} - {t_1}) + ({t_2} - {t_1}) = 0 ⇒ t 1 t 2 ( t 2 − t 1 ) + ( t 2 − t 1 ) = 0
⇒ ( t 2 − t 1 ) ( t 1 t 2 + 1 ) = 0 \Rightarrow ({t_2} - {t_1})({t_1}{t_2} + 1) = 0 ⇒ ( t 2 − t 1 ) ( t 1 t 2 + 1 ) = 0
As t 2 − t 1 ≠ 0 {t_2} - {t_1} \ne 0 t 2 − t 1 = 0
∴ \therefore ∴ t 1 t 2 + 1 = 0 {t_1}{t_2} + 1 = 0 t 1 t 2 + 1 = 0
t 1 t 2 = − 1 {t_1}{t_2} = - 1 t 1 t 2 = − 1
Now, lenght of PQ
= ( t 1 2 − t 2 2 ) 2 + ( 2 t 1 − 2 t 2 ) 2 = \sqrt {{{\left( {t_1^2 - t_2^2} \right)}^2} + {{\left( {2{t_1} - 2{t_2}} \right)}^2}} = ( t 1 2 − t 2 2 ) 2 + ( 2 t 1 − 2 t 2 ) 2
= ( t 1 + t 2 ) 2 ( t 1 − t 2 ) 2 + 4 ( t 1 − t 2 ) 2 = \sqrt {{{\left( {{t_1} + {t_2}} \right)}^2}{{\left( {{t_1} - {t_2}} \right)}^2} + 4{{\left( {{t_1} - {t_2}} \right)}^2}} = ( t 1 + t 2 ) 2 ( t 1 − t 2 ) 2 + 4 ( t 1 − t 2 ) 2
= ( t 1 − t 2 ) ( t 1 + t 2 ) 2 + 4 = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} = ( t 1 − t 2 ) ( t 1 + t 2 ) 2 + 4
= ( t 1 − t 2 ) t 1 2 + t 2 2 + 2 t 1 t 2 + 4 = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2{t_1}{t_2} + 4} = ( t 1 − t 2 ) t 1 2 + t 2 2 + 2 t 1 t 2 + 4
= ( t 1 − t 2 ) t 1 2 + t 2 2 + 2 ( − 1 ) + 4 = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2( - 1) + 4} = ( t 1 − t 2 ) t 1 2 + t 2 2 + 2 ( − 1 ) + 4
= ( t 1 − t 2 ) t 1 2 + t 2 2 + 2 = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2} = ( t 1 − t 2 ) t 1 2 + t 2 2 + 2
= ( t 1 − t 2 ) t 1 2 + t 2 2 − 2 ( − 1 ) = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2( - 1)} = ( t 1 − t 2 ) t 1 2 + t 2 2 − 2 ( − 1 )
= ( t 1 − t 2 ) t 1 2 + t 2 2 − 2 t 1 t 2 = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2{t_1}{t_2}} = ( t 1 − t 2 ) t 1 2 + t 2 2 − 2 t 1 t 2
= ( t 1 − t 2 ) ( t 1 − t 2 ) 2 = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} - {t_2}} \right)}^2}} = ( t 1 − t 2 ) ( t 1 − t 2 ) 2
= ( t 1 − t 2 ) 2 = {\left( {{t_1} - {t_2}} \right)^2} = ( t 1 − t 2 ) 2
Given, length of P Q = ( t 1 − t 2 ) 2 = 6.25 PQ = {\left( {{t_1} - {t_2}} \right)^2} = 6.25 PQ = ( t 1 − t 2 ) 2 = 6.25
⇒ t 1 − t 2 = 2.5 \Rightarrow {t_1} - {t_2} = 2.5 ⇒ t 1 − t 2 = 2.5
Now, Area of Δ O P Q \Delta OPQ Δ OPQ
= ∣ 1 2 ( 2 t 2 t 1 2 − 2 t 1 t 2 2 ) ∣ = \left| {{1 \over 2}\left( {2{t_2}t_1^2 - 2{t_1}t_2^2} \right)} \right| = 2 1 ( 2 t 2 t 1 2 − 2 t 1 t 2 2 )
= ∣ 1 2 × 2 t 1 t 2 ( t 1 − t 2 ) ∣ = \left| {{1 \over 2} \times 2{t_1}{t_2}\left( {{t_1} - {t_2}} \right)} \right| = 2 1 × 2 t 1 t 2 ( t 1 − t 2 )
= ∣ t 1 t 2 × ( t 1 − t 2 ) ∣ = \left| {{t_1}{t_2} \times \left( {{t_1} - {t_2}} \right)} \right| = ∣ t 1 t 2 × ( t 1 − t 2 ) ∣
= ∣ − 1 × 2.5 ∣ = \left| { - 1 \times 2.5} \right| = ∣ − 1 × 2.5 ∣
= 2.5 = 2.5 = 2.5
∴ \therefore ∴ 10 Δ \Delta Δ OPQ =
= 10 × 25 10 = 25 = 10 \times {{25} \over {10}} = 25 = 10 × 10 25 = 25
