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One extremity of a focal chord of y^2 = ...

One extremity of a focal chord of `y^2 = 16x` is `A(1,4)`. Then the length of the focal chord at A is

A

22

B

25

C

24

D

20

Text Solution

Verified by Experts

The correct Answer is:
B
(i) First find the focus of given parabola
(ii) Then, find the slope of the focal chord by using `m = (y_(2) - y_(1))/(x_(2) - x_(1))`
(iii) Now, find the length of the focal chord by the using the formula `4 a sec^(2) alpha`.
Equation of given parabola is `y^(2) = 16x` its formula is (4,0) since, slope of line pasing through `(x_(1), y_(1))` and `(x_(2), y_(2))` is given by `m = tan theta = (y_(2) - y_(1))/(x_(2) - x_(2))`
`:.` Slope of focal chord having one end point is (1,4) is
`m = tan alpha = (4 -0)/(1 - 4) = - (4)/(3)`
Where `'alpha'` is the inclination of focal with X-axis.
since the length of focal chord `= 4a cosec^(2) alpha`
`:.` The required length of the focal chord
`= 16 [1 + cot^(2) alpha] [:' a = 4 "and" cosec^(2) alpha = 1 + cot^(2) alpha]`
`= 16 [1 + (9)/(16)] = 25` units `[:' cot alpha = (1)/(tan alpha) = - (3)/(4)]`
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Knowledge Check

  • If the lsope of the focal chord of y^(2)=16x is 2, then the length of the chord, is

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