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Question: The length of a focal chord of the parabola \[{y^2} = 4ax\] at a distance b from the vertex is c, th...

The length of a focal chord of the parabola y2=4ax{y^2} = 4ax at a distance b from the vertex is c, then:
A.2a2=bc2{a^2} = bc
B. a3=b2c{a^3} = {b^2}c
C. ac=b2ac = {b^2}
D. b2c=4a3{b^2}c = 4{a^3}

Explanation

Solution

We draw a rough figure for the situation where we have a parabola and a focal chord of length c. Using the property of vertically opposite angles we have the angle made by the chord with x-axis equal on both sides of the axis. With the help or right triangle formed by the shortest distance from the vertex to the chord we find cosecant of the angle. Substitute the value of cosecant of the angle in the formula for length of a chord.

  • Focal chord is a chord that passes through the focus of the parabola
  • In a right triangle having angleθ\theta , sinθ=\sin \theta = perpendicularly divided by hypotenuse and cosecθ=1sinθ\cos ec\theta = \dfrac{1}{{\sin \theta }}.
    *Length of focal chord of parabola y2=4ax{y^2} = 4ax making an angle θ\theta with the x-axis is given by 4acosec2θ4a\cos e{c^2}\theta

Complete step-by-step answer:

We are given a parabola y2=4ax{y^2} = 4ax
Let us assume that the chord cuts the X-axis at point D(a,0)
Then according to the question we are given the shortest distance from center to the chord is b.
Length of the focal chord is c.
The distance OD=aOD = a.
Let us assume the focal chord makes an angle x with the X-axis.

Since, we know vertically opposite angles are equal. Then, ODC=x\angle ODC = x.
Since, the shortest distance is the perpendicular distance, then OCD=90\angle OCD = {90^ \circ }.
In right triangle OCD,
sinx=OCOD\sin x = \dfrac{{OC}}{{OD}}
Substitute the value of OD as a and OC as b
sinx=ba\Rightarrow \sin x = \dfrac{b}{a}
Since we know cosecx=1sinx\cos ecx = \dfrac{1}{{\sin x}}
cosecx=1ba\Rightarrow \cos ecx = \dfrac{1}{{\dfrac{b}{a}}}
cosecx=ab\Rightarrow \cos ecx = \dfrac{a}{b} … (1)
We know focal chord of a parabola y2=4ax{y^2} = 4axmaking an angle θ\theta with the X-axis is of length 4acosec2θ4a\cos e{c^2}\theta
Therefore, we can write c=4acosec2xc = 4a\cos e{c^2}x
Substitute the value of cosecx\cos ecxfrom equation (1)
c=4a(ab)2\Rightarrow c = 4a{\left( {\dfrac{a}{b}} \right)^2}
Open the value on the bracket.
c=4a3b2\Rightarrow c = \dfrac{{4{a^3}}}{{{b^2}}}
Cross multiply the value in denominator of RHS to LHS
b2c=4a3\Rightarrow {b^2}c = 4{a^3}

Therefore, option D is correct.

Note: Students are likely to make the mistake of assuming the shortest distance as OD, which makes b equal to a. Keep in mind the shortest distance is the perpendicular distance and here the chord is not perpendicular to the x-axis so the perpendicular distance from vertex cannot be a.