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Question: If two distinct chords of parabola \[{{y}^{2}}=4ax\], passing through \[\left( a,2a \right)\] are bi...

If two distinct chords of parabola y2=4ax{{y}^{2}}=4ax, passing through (a,2a)\left( a,2a \right) are bisected on the line x+y=1x+y=1, then the length of latus rectum is
(a) 2
(b) 1
(c) 4
(d) 5

Explanation

Solution

The figure showing the given data is

We solve this problem by assuming the mid - point as (h,k)\left( h,k \right) that lies on the line x+y=1x+y=1. Then we use the formula of the line equation having the mid – point of a curve SSthat is S1=S11{{S}_{1}}={{S}_{11}}. If Sy24ax=0S\equiv {{y}^{2}}-4ax=0 then the representation S1,S11{{S}_{1}},{{S}_{11}} with respect to point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) is given as
S1=yy12a(x+x1)\Rightarrow {{S}_{1}}=y{{y}_{1}}-2a\left( x+{{x}_{1}} \right)
S11=y124ax1\Rightarrow {{S}_{11}}={{y}_{1}}^{2}-4a{{x}_{1}}
By using the above results we find the range of a'a' to get the latus rectum as 4a'4a'

Complete step-by-step solution:
Let us assume that the point at which the chords are bisected as (h,k)\left( h,k \right)
We are given that the chords are bisected on the line x+y=1x+y=1
We know that the point (h,k)\left( h,k \right) satisfies the equation x+y=1x+y=1
Now, by substituting the point in the line we get

& \Rightarrow h+k=1 \\\ & \Rightarrow k=1-h \\\ \end{aligned}$$ So, we can take the mid – point as $$\left( h,1-h \right)$$ We know that the formula of line equation having mid – point of a curve $$S$$ that is $${{S}_{1}}={{S}_{11}}$$. If $$S\equiv {{y}^{2}}-4ax=0$$ then the representation $${{S}_{1}},{{S}_{11}}$$ with respect to point $$\left( {{x}_{1}},{{y}_{1}} \right)$$ is given as $$\Rightarrow {{S}_{1}}=y{{y}_{1}}-2a\left( x+{{x}_{1}} \right)$$ $$\Rightarrow {{S}_{11}}={{y}_{1}}^{2}-4a{{x}_{1}}$$ Now, by using the mid – point $$\left( h,1-h \right)$$ to the curve $$S\equiv {{y}^{2}}-4ax=0$$ we get the equation of chord as $$\begin{aligned} & \Rightarrow {{S}_{1}}={{S}_{11}} \\\ & \Rightarrow y\left( 1-h \right)-2a\left( x+h \right)={{\left( 1-h \right)}^{2}}-4ah \\\ & \Rightarrow y\left( 1-h \right)-2ax={{\left( 1-h \right)}^{2}}-2ah \\\ \end{aligned}$$ We are given that the chords are drawn from the point $$\left( a,2a \right)$$ so, we can say that the point $$\left( a,2a \right)$$ satisfies the above equation. By substituting the point $$\left( a,2a \right)$$ in above equation we get $$\begin{aligned} & \Rightarrow 2a\left( 1-h \right)-2a\left( a \right)={{\left( 1-h \right)}^{2}}-2ah \\\ & \Rightarrow 2a-2{{a}^{2}}={{\left( 1-h \right)}^{2}} \\\ \end{aligned}$$ We know that the square of any number is greater than 0. So, we can write $$\Rightarrow {{\left( 1-h \right)}^{2}}>0$$ By substituting the required equation in above equation we get $$\Rightarrow 2a-2{{a}^{2}}>0$$ Now, by multiplying with negative sign the inequality changes as $$\begin{aligned} & \Rightarrow 2{{a}^{2}}-2a<0 \\\ & \Rightarrow 2a\left( a-1 \right)<0 \\\ & \Rightarrow a\in \left( 0,1 \right) \\\ \end{aligned}$$ We know that the length of latus rectum of given parabola is $$'4a'$$ Therefore the range of latus rectum can be taken as $$\Rightarrow 4a\in \left( 0,4 \right)$$ Here, we can say that the length of latus rectum can take any values in the domain $$\left( 0,4 \right)$$ **Therefore, the possible values of latus rectum according to options are 1, 2. So, option (a) and option (b) are correct answers.** **Note:** Students may make mistakes in solving the inequality. While multiplying with a negative sign the inequality changes. Here, we have $$\Rightarrow 2a-2{{a}^{2}}>0$$ After multiplying with a negative sign the equation changes as $$\Rightarrow 2{{a}^{2}}-2a<0$$ But students miss this point and do not change the inequality which results in wrong answers. Also, we have the range of latus rectum as $$\Rightarrow 4a\in \left( 0,4 \right)$$ This means the latus rectum can have values in the domain $$\left( 0,4 \right)$$ but will not equal to ‘0’ or ‘4’. Students may miss this point and give the answer ‘4’ also as the correct answer. This will be wrong.