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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
- A \({\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}\)
- B \(\;{\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} - 2{y^2}\)
- C \(\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} + 2{y^2}\)
- D \(\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} - 2{y^2}\)
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:-…
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