The locus of the feet of the perpendicular to any tangent of an ellipse from the foci is

Options

A. `x^(2) + y^(2) = b^(2)`
B. `x^(2) + y^(2) = a^(2)`
C. `x^(2) + y^(2) = a^(2) + b^(2)`
D. none of these

Answer

B

Explanation

Let (h,k) be the foot of the perpendicular drawn from the focus S (ae, 0) on tangent
`y = mx + sqrt(a^(2)m^(2) + b^(2))" "…(i)`
at point P on the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`. Then,
`k = mh + sqrt(a^(2)m^(2) + b^(2)) rArr (k - mh)^(2) = a^(2)m^(2) + b^(2) " "...(ii)`
Also, SP is perpendicular to (i).
`therefore m xx (k - 0)/(h - ae) = -1 rArr h + mk = ae " "...(iii)`
Form (ii) and (iii), we get
`(k - mh)^(2) + (h + mk)^(2) = a^(2)m^(2) + b^(2) + a^(2) e^(2)`
`rArr (h^(2) + k^(2)) (l + m)^(2) = a^(2) (1 + m^(2)) rArr h^(2) + k^(2) = a^(2)`
Hence, the locus of (h, k) is `x^(2) + y^(2) = a^(2)`.