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Question: If \(b\) and \(c\) are lengths of the segments of the focal chord of a parabola\[{y^2} = 4ax\], then...

If bb and cc are lengths of the segments of the focal chord of a parabolay2=4ax{y^2} = 4ax, then the length of the latus rectum is
A. bc(b+c)\dfrac{{bc}}{{(b + c)}}
B. bc\sqrt {bc}
C. (b+c)2\dfrac{{(b + c)}}{2}
D. 2bc(b+c)\dfrac{{2bc}}{{(b + c)}}

Explanation

Solution

This question is based on the properties of the curve parabola.Concepts like parametric point, focus, and focal chords are used to solve this question. The basics about focal chords must be clear, and the formula should be remembered.

Complete step by step answer:
The chord of the parabola which passes through the focus is called the focal chord.And chord to y2=4ax{y^2} = 4ax which passes through the focus is called a focal chord of the parabola y2=4ax{y^2} = 4ax.

Let y2=4ax{y^2} = 4ax be the equation of a parabola and (at12,2at1)(at_1^2,2a{t_1}) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at22,2at2)(at_2^2,2a{t_2})
Then, PS and SQ, where S is the focus (a,0)(a,0), have the same slopes
⇒2at1−0at12−a=2at2−0at22−a\Rightarrow \dfrac{{2a{t_1} - 0}}{{a{t_1}^2 - a}} = \dfrac{{2a{t_2} - 0}}{{a{t_2}^2 - a}}

Simplifying the equation and cross-multiplying gives
⇒t1t22−t1=t2t12−t2\Rightarrow {t_1}{t_2}^2 - {t_1} = {t_2}{t_1}^2 - {t_2}
⇒t1t2(t2−t1)=−(t2−t1)\Rightarrow {t_1}{t_2}({t_2} - {t_1}) = - ({t_2} - {t_1})
So, we have a relation between the focal points
⇒t1t2=−1\Rightarrow {t_1}{t_2} = - 1
Hence, t2=−1t1{t_2} = \dfrac{{ - 1}}{{{t_1}}}, i.e., the point Q is (at12,−2at1)(\dfrac{a}{{{t_1}^2}}, - \dfrac{{2a}}{{{t_1}}}).
The extremities of a focal chord of the parabola y2=4ax{y^2} = 4axmay be taken as the points t1{t_1} and −1t1\dfrac{{ - 1}}{{{t_1}}}.

Now, calculating the length of the segment of the focal chord as shown below
⇒l1=(at12−a)2+(2at1−0)2\Rightarrow {l_1} = \sqrt {{{(at_1^2 - a)}^2} + {{(2a{t_1} - 0)}^2}}
⇒l1=at12+a\Rightarrow {l_1} = at_1^2 + a
Now, calculating the length of the segment of the focal chord as shown below,
⇒l2=(at12−a)2+(−2at1−0)2\Rightarrow {l_2} = \sqrt {{{(\dfrac{a}{{{t_1}^2}} - a)}^2} + {{( - \dfrac{{2a}}{{{t_1}}} - 0)}^2}}
⇒l2=(at12+a)\Rightarrow {l_2} = (\dfrac{a}{{{t_1}^2}} + a)
Now, calculating the length of the focal chord
⇒l=(at12−at12)2+(2at1−(−2at1))2\Rightarrow l = \sqrt {{{(at_1^2 - \dfrac{a}{{{t_1}^2}})}^2} + {{(2a{t_1} - ( - \dfrac{{2a}}{{{t_1}}}))}^2}}
⇒l=a(t1+1t1)2\Rightarrow l = a{({t_1} + \dfrac{1}{{{t_1}}})^2}

Now, deriving a relation between the length of the focal chord and the segments of the focal chord
⇒l=4l1l2(l1+l2)\Rightarrow l = \dfrac{{4{l_1}{l_2}}}{{({l_1} + {l_2})}}
Thus, the length of the semi-major axis can be given as,
⇒s=12l\Rightarrow s = \dfrac{1}{2}l
Or the length of the semi-major axis is the harmonic mean of the segments of the focal chord. Thus, if bb and cc are the lengths of the segments of the focal chord, then the length of the semi-major axis is given by:
∴s=2bc(b+c)\therefore s = \dfrac{{2bc}}{{(b + c)}}

Thus, option D is the correct answer.

Note: Latus rectum of a parabola is a line segment passing through its focus and perpendicular to its axis. Length of the latus rectum of the parabola is always equal to four times its focal length or the distance between the vertex of the parabola and focus point. The relation between the length of the focal chord and its segment can be used as a direct result in a variety of other questions.