The locus of the point of intersection of tangents to the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`, which make complementary angles with x - axis, is

Options

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

Answer

D

Explanation

The equation of any tangent to the ellipse
`x^(2)/a^(2) + y^(2)/b^(2) = 1` is
`y = mx pm sqrt(a^(2)m^(2) + b^(2))" "…(i)`
Let P (h, k) be the point of intersection of tangents. If (i) passes through P (h, k), then
`k = mh pm sqrt(a^(2)m^(2) + b^(2))`
`rArr (k - mh)^(2) = a^(2)m^(2) + b^(2) rArr m^(2)(h^(2) - a^(2)) - 2mhk + k^(2) - b^(2) = 0`
This gives two values of m, say `m_(1) and m_(2)`. These values represent the slopes of the tangents passing through P.
If the tangents drawn from P make complementary angles with x - axis, then
`m_(1)m_(2) = 1 rArr (k^(2) - b^(2))/(h^(2) - a^(2)) = 1 rArr h^(2) - k^(2) = a^(2) - b^(2)`
Hence, the locus of (h, k) is `x^(2) - y^(2) = a^(2) - b^(2)`