ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

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

  1. A \({\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}\)
  2. B \(\;{\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} - 2{y^2}\)
  3. C \(\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} + 2{y^2}\)
  4. D \(\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} - 2{y^2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \({\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}\)

Step-by-step Solution

Detailed explanation

\(x^{2}+3 y^{2}=6\) \(\Rightarrow \frac{x^{2}}{6}+\frac{y^{2}}{2}=1\) Now, we know that any tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is given by: \(y=m x+\sqrt{a^{2} m^{2}+b^{2}}\) So, the equation of the tangent to the given ellipse is:-…