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Question: If b, k are the intercept of a focal chord of the parabola \(y ^ { 2 } = 4 a x\), then K is equal to...

If b, k are the intercept of a focal chord of the parabola y2=4axy ^ { 2 } = 4 a x, then K is equal to

A

abb−a\frac { a b } { b - a }

B

bb−a\frac { b } { b - a }

C

ab−a\frac { a } { b - a }

D

aba−b\frac { a b } { a - b }

Answer

abb−a\frac { a b } { b - a }

Explanation

Solution

Let be the ends of focal chords

∴ t1t2=−1t _ { 1 } t _ { 2 } = - 1 . If S is the focus and P, Q are the ends of the focal chord, then

=a(t12+1)=b= a \left( t _ { 1 } ^ { 2 } + 1 \right) = b (Given).... (i)

∴ =a(1t12+1)= a \left( \frac { 1 } { t _ { 1 } ^ { 2 } } + 1 \right) (Given) [∵t2=−1t1⇒t22=1t12]\left[ \because t _ { 2 } = - \frac { 1 } { t _ { 1 } } \Rightarrow t _ { 2 } ^ { 2 } = \frac { 1 } { t _ { 1 } ^ { 2 } } \right]

=a(t12+1)t12=k= \frac { a \left( t _ { 1 } ^ { 2 } + 1 \right) } { t _ { 1 } ^ { 2 } } = k ....(ii), ∴ bk=t12\frac { b } { k } = t _ { 1 } ^ { 2 } [Divide (i) by (ii)]

Putting in (1), we get a(bk+1)=b⇒abk+a=ba \left( \frac { b } { k } + 1 \right) = b \Rightarrow \frac { a b } { k } + a = b⇒ k=abb−ak = \frac { a b } { b - a }`