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  • Let     be a variable point on the parabola   <img alt="\mathrm {y=4 x^2+1.}" src

Let   \mathrm P  be a variable point on the parabola   \mathrm {y=4 x^2+1.}  Then, the locus of the mid-point of the point  \mathrm {P} and the foot of the perpendicular drawn from the point \mathrm {P}  to the line \mathrm {y=x \: \: is\: a(3 x-y)^2+(x-3 y)+a=0}. Then the value of \mathrm {a}  is_______

Option: 1

3


Option: 2

2


Option: 3

1


Option: 4

0


Answers (1)

best_answer

Let  \mathrm {R(h, k) }  be the mid point of the point \mathrm {P } and the foot of the perpendicular drawn from \mathrm {P } to the line  \mathrm {y=x. \: \: Let Q(t, t) } is the foot of perpendicular.

\mathrm {Then, \frac{k-t}{h-t}=-1}

\mathrm {\Rightarrow t=\frac{h+k}{2}} \mathrm {Also, R(h, k)=\left(\frac{x+t}{2}, \frac{y+t}{2}\right)}

\begin{aligned} & \mathrm {\Rightarrow \quad R(h, k)=\left(\frac{x}{2}+\frac{h}{4}+\frac{k}{4}, \frac{y}{2}+\frac{h}{4}+\frac{k}{4}\right) }\\ & \mathrm { \therefore \quad x=\frac{3 h}{2}-\frac{k}{2} \text { and } y=\frac{3 k}{2}-\frac{h}{2}} \end{aligned}

Now, put \mathrm {x \: \: \&\: \: y} in  \mathrm {y=4 x^2+1} , we get

\begin{aligned} & \mathrm {\frac{3 k-h}{2}=4\left(\frac{3 h-k}{2}\right)^2+1 }\\ \Rightarrow & \mathrm {\frac{3 k-h}{2}=\frac{4(3 h-k)^2}{4}+1 }\Rightarrow \mathrm {\frac{3 k-h}{2}=(3 h-k)^2+1 }\\ \Rightarrow & \mathrm {(3 k-h)=2(3 h-k)^2+2 }\\ \Rightarrow & \mathrm {2(3 h-k)^2-(3 k-h)+2=0 }\\ \Rightarrow & \mathrm {2(3 h-k)^2+(h-3 k)+2=0 }\\ \therefore & \text { The required locus is } \mathrm {2(3 x-y)^2+(x-3 y)+2=0} . \end{aligned}
 

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Anam Khan

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