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  • The locus of the mid-points of the normal chords of the parabola x2 = 4ay isOption: 1 <img alt="x^{4}-2 a(y-2 a) x^{2}+8 a^{4}=0

The locus of the mid-points of the normal chords of the parabola x2 = 4ay is

Option: 1

x^{4}-2 a(y-2 a) x^{2}+8 a^{4}=0


Option: 2

x^{4}-2 a(y-2 a) +8 a^{4}=0


Option: 3

y^{4}-2 a(x-2 a) y^{2}+8 a^{4}=0


Option: 4

y^{4}-2 a(x-2 a) +8 a^{4}=0


Answers (1)

best_answer

The equation of a parabola is x^2=4ay

Equation of the normal at any point (2at,at2) of the parabola

x+t y=2 a t+a t^{3} \quad....(i)

Let PQ be the normal, whose mid-point is M(h, k). Therefore, equation of PQ is

\begin{aligned} & T=S_{1} \\ \Rightarrow & xh-2 a(y+k)=h^{2}-4 a k \\ \Rightarrow & hx-2 ay-h^{2}+2 a k=0 \quad....(ii)\end{aligned}

Equations (i) and (ii) are identical if

\frac{1}{h}=\frac{t}{-2 a}=\frac{2 a t+a t^{3}}{h^{2}-2 a k}

\Rightarrow \quad t=-\frac{2 a}{h} \text { and } \frac{1}{h}=\frac{2 a t+a t^{3}}{h^{2}-2 ak}

Eliminating t

\frac{1}{h}=\frac{2 a\left(\frac{-2 a}{h}\right)+a\left(\frac{-2 a}{h}\right)^{3}}{h^{2}-2 a k}

\begin{array}{l}{\Rightarrow \quad\left(h^{2}-2 a k\right)=-2 a\left(2 a+a\left(\frac{-2 a}{h}\right)^{2}\right)} \\ {\Rightarrow \quad h^{2}\left(h^{2}-2 a k\right)=-2 a\left(2 a h^{2}+4 a^{3}\right)} \\ {\Rightarrow \quad h^{2}\left(h^{2}-2 a k\right)=-4 a^{2} h^{2}-8 a^{4}} \\ {\Rightarrow \quad h^{4}-2 a(k-2 a) h^{2}+8 a^{4}=0}\end{array}

Hence the locus of M (h,k) is

x^{4}-2 a(y-2 a) x^{2}+8 a^{4}=0

Posted by

seema garhwal

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