JEE MAIN - Mathematics (2014 (Offline) - No. 12)
The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is :
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$
Explanation
Given $$e{q^n}$$ of ellipse can be written as
$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$
Now, equation of any variable tangent is
$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$
where $$m$$ is slope of the tangent
So, equation of perpendicular line drawn-
from center to tangent is
$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Eliminating $$m,$$ we get
$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$
$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$
$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$
Now, equation of any variable tangent is
$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$
where $$m$$ is slope of the tangent
So, equation of perpendicular line drawn-
from center to tangent is
$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Eliminating $$m,$$ we get
$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$
$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$
$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$
Comments (0)
