The sum fo the squares of the perpendicular on any tangent to the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` from two points on the mirror axis, each at a distance `sqrt(a^(2) - b^(2))` from the centre, is

Options

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

Answer

A

Explanation

Let e be the eccentricity of the ellipse. Then,
`b^(2) = a^(2)(1 - e^(2)) rArr a^(2) - b^(2) = a^(2)e^(2) rArr sqrt(a^(2) - b^(2)) = ae.`
Thus, two points on the minor axis are
P (0, ae) and Q (0, ae).
The equation of any tangent to the given ellipse is
`y = mx + sqrt(a^(2)m^(2) + b^(2)) rArr mx - y + sqrt(a^(2)m^(2) + b^(2)) = 0 ...(i)`
Now,
`p_(1)` = Length of the `bot` from P (0, ae) on (i)
`= (-ae + sqrt(a^(2)m^(2) + b^(2)))/(sqrt(1 + m^(2)))`
and, `P_(2) = ` Length of the `bot` from Q (0, - ae) on (i)
`= (ae + sqrt(a^(2)m^(2) + b^(2)))/(sqrt(1 + m^(2)))`
`therefore p_(1)^(2) + p_(2)^(2) = (2(a^(2)e^(2) + a^(2)m^(2) + b^(2)))/(1 + m^(2))`
`rArr p_(1)^(2) + p_(2)^(2) = (2{a^(2)e^(2) + a^(2)m^(2) + a^(2)(1 - e^(2))})/(1 + m^(2)) = 2a^(2)`