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Question: The curve for which the normal at any point \[\left( {x,y} \right)\] and the line joining the origin...

The curve for which the normal at any point (x,y)\left( {x,y} \right) and the line joining the origin to the points from the isosceles triangle with xx–axis as a base, is
A. An ellipse
B. A rectangular hyperbola
C. A circle
D. None of the above

Explanation

Solution

First we will first take the OPG is an isosceles triangle, OM = MG = subnormal{\text{OM = MG = subnormal}}. Taking the equation x=ydydxx = y\dfrac{{dy}}{{dx}} by separating variables and then integrate it to find the equation for the final answer.

Complete step by step answer:

We are given that the curve for which the normal at any point (x,y)\left( {x,y} \right) and the line joining the origin to the points from the isosceles triangle with xx–axis as a base.

It is given that the OPG is an isosceles triangle.

Therefore OM = MG = subnormal{\text{OM = MG = subnormal}}.

x=ydydx xdx=ydy  \Rightarrow x = y\dfrac{{dy}}{{dx}} \\\ \Rightarrow xdx = ydy \\\

Integrating the above equation on both sides, we get
x22=y22+c\Rightarrow \dfrac{{{x^2}}}{2} = \dfrac{{{y^2}}}{2} + c
Multiplying the above equation by 3 on both sides, we get

x2=y2+2c x2y2=2c  \Rightarrow {x^2} = {y^2} + 2c \\\ \Rightarrow {x^2} - {y^2} = 2c \\\

Taking a=2ca = 2c in the above equation, we get
x2y2=a\Rightarrow {x^2} - {y^2} = a
We know that the standard equation of the rectangular hyperbola (xx0)2+(yy0)2=a2{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {a^2}, where (x0,y0)\left( {{x_0},{y_0}} \right) is the point of intersection of hyperbola and aa is the length.
Compare the above equation with the standard equation of a rectangular hyperbola.
This implies that it is a rectangular hyperbola.
Hence, option B is correct.

Note: In solving these types of questions, the key concept is to know that when a triangle is an isosceles triangle, then the midlines are subnormal. Then we can easily find the equation of the curve to find the final answer.