Question

The length of the normal chord to the parabola y^2=4x, which subtends a right angle at the vertex is

Answer 1

Answer:

(6) x (3)^(1/2)

Step-by-step explanation:

Let PQ is the common chord.

At any point coorinates of P and Q are (t12,2t1) and (t22,2t2)

As the chord subtends an angle of 90∘, relation between t1 and t2 will be,

t2=−t1−2t1 →Eq(1)

If O is the origin, then OP and OQ are perpendicular to each other. In that case, both their slopes multiplication will be -1.

Thus,

2t1t12∗2t2t22=−1

⇒t1.t2=−4→Eq(2)

Putting values of t2 from Eq(1)

⇒t1(−t1−2t1)=−4

⇒−t13−2t1=−4t1

⇒t13−2t1=0

Solving, above, we get, t1=0 or t1=(2)^(1/2)

As, t1=0 is not possiblee, so t1=(2)^(1/2)

Putting value of t1 in Eq(2), we get, t2=−(22)^(1/2)

So, coordinates of chord areP(2,22–√) and Q(8,−(42)^(1/2))

So, length of PQ will be (62–√)2+62−−−−−−−−−−√)=(108)^(1/2)=(6)*(3)^(1/2)