Find the locus of the foot of the perpendicular drawn from the center upon any tangent to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1.`

Options

A. `(x^(2) - y^(2))^(2) = a^(2)x^(2) + b^(2)y^(2)`
B. `(x^(2) - y^(2))^(2) = a^(2)x^(2) - b^(2)y^(2)`
C. `(x^(2) + y^(2))^(2) = a^(2)x^(2) + b^(2)y^(2)`
D. `(x^(2) + b^(2))^(2) = a^(2)x^(2) - b^(2)y^(2)`

Answer

C

Explanation

Let P (h, k) be the foot of the perpendicular drawn from the centre C (0, 0) of the ellipse to any tangent
`y = mx + sqrt(a^(2)m^(2) + b^(2))" "…(i)`
Then, `k = mh + sqrt(a^(2)m^(2) + b^(2)) rArr (k - mh)^(2) = a^(2)m^(2) + b^(2) " "…(ii)`
Since CP is perpendicular to the tangent given in (i).
`therefore k/h xx m = -1 rArr m = - h/k`
Substituting the value of m in (ii), we get
`(h^(2) + k^(2))^(2) = a^(2)h^(2) + b^(2)k^(2)`
Hence, the locus of P (h, k) is `a^(2)x^(2) + b^(2)y^(2) = (x^(2) + y^(2))^(2)`