JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q80

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MathematicsSingle correct+4 / -1Question 80 of 90

The locus of the foot of perpendicular drawn from the centre of the ellipse $x^2+3y^2=6$ on any tangent to it is :

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

\left(x^2+y^2\right)^2=6x^2-2y^2

\left(x^2-y^2\right)^2=6x^2+2y^2

\left(x^2-y^2\right)^2=6x^2-2y^2

Answer and worked solution

Official answer: A

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , the foot $(h,k)$ of the perpendicular from the centre lies on the tangent $hx+ky=h^2+k^2$ . That tangent condition gives the locus $(x^2+y^2)^2=a^2x^2+b^2y^2$ .

Here $a^2=6$ , $b^2=2$ :
$\left(x^2+y^2\right)^2=6x^2+2y^2.$

Topics:Ellipse Tangents and normals to conics Locus

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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 80