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  • Two equal parabola’s have the same focus and their axes are at right angles, a normal to one is perpendicular to a normal to the other. The locus of the point of in

Two equal parabola’s have the same focus and their axes are at right angles, a normal to one is perpendicular to a normal to the other. The locus of the point of intersection  of these normal is a : 

 

Option: 1

straight line
 


Option: 2

 Circle 


Option: 3

Parabola
 


Option: 4

Ellipse


Answers (1)

best_answer

Taking the common focus as origin and axis of the parabola as axis of the co-ordinate, the equation of the parabola may be written as 
\mathrm{\begin{aligned} & y^2=4 a(x+a)...........(1) \\& x^2=4 a(y+a)........(2) \end{aligned} }
where \mathrm{4 a }  is the latus rectum of each. Equation to any normal to (1) is 


\mathrm{y=m(x+a)-2 a m-a m^3 }.............(3)
Equation to any normal to (2) is
\mathrm{x=m^{\prime}(y+a)-2 a m^{\prime}-a m^{\prime 3} }...........(4)
But the two normals are at right angles then,
\mathrm{\mathrm{m} \times \frac{1}{\mathrm{~m}^{\prime}}=-1 \text { or } \mathrm{m}^{\prime}=-\mathrm{m} }
Substituting \mathrm{\mathrm{m}^{\prime}=-\mathrm{m} } in equation (4)
\mathrm{x=-m(y+a)+2 a m+a m^3 }..........(5)
The locus of the point of intersection of the normals will be the elimination of \mathrm{\mathrm{m} } from (3) and (5). So adding (3) and (5) after taking the terms to R.H.S. in both cases we get,
\mathrm{\begin{aligned} & O=m x-y-x-m y \\ & x+y=m(x-y) \text { or } m=\frac{x+y}{x-y} \end{aligned} }
Putting the value of \mathrm{ \mathrm{m} } in (3) we get,
\mathrm{\begin{array}{ll} y=\frac{x+y}{x-y}(x+a)-2 a \frac{x+y}{x-y}-a\left(\frac{x+y}{x-y}\right)^3 \\ y(x-y)^3=(x+y)\left[(x+a)(x-y)^2-2 a(x-y)^2-a(x+y)^2\right] \\ =(x+y)\left[(x-y)^2(x+a-2 a)-a(x+y)^2\right] \\ =(x+y)\left[x(x-y)^2-2 a\left(x^2+y^2\right)\right] \\ \Rightarrow \quad 2 a\left(x^2+y^2\right)(x+y)=(x+y) x(x-y)^2-y(x-y)^3 \\ \Rightarrow \quad 2 a\left(x^2+y^2\right)(x+y)=(x-y)^2\left[x^2+x y-x y+y^2\right] \\ \Rightarrow \quad 2 a(x+y)=(x-y)^2 \end{array} }
This is the required locus which is a parabola.

 

Posted by

Gaurav

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