Correct Answer - Option 2 :
5√2x + 4√2y = 40
CONCEPT:
Equation of tangent to the horizontal ellipse \(\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) where 0 < b < a at the point (x1, y1) is given by: \(\frac{{x\; \cdot {x_1}}}{{{a^2}}} + \frac{{y\; \cdot {y_1}}}{{{b^2}}} = 1\)
Equation of tangent to the vertical ellipse\(\;\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\) where 0 < b < a at the point (x1, y1) is given by: \(\frac{{x\; \cdot {x_1}}}{{{b^2}}} + \frac{{y\; \cdot {y_1}}}{{{a^2}}} = 1\)
CALCULATION:
Given: Equation of ellipse is 25x2 + 16y2 = 400
here, we have to find the equation of tangent to the given ellipse at the point \(P (2\sqrt 2, \frac{5}{\sqrt 2})\)
The given equation of ellipse can be re-written as:\(\;\frac{{{x^2}}}{{{16}}} + \frac{{{y^2}}}{{{25}}} = 1\)
As we can see that, the given ellipse is a vertical ellipse
As we know that, equation of tangent to the vertical ellipse\(\;\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1\) where 0 < b < a at the point (x1, y1) is given by: \(\frac{{x\; \cdot {x_1}}}{{{b^2}}} + \frac{{y\; \cdot {y_1}}}{{{a^2}}} = 1\)
Here, \(x_1 = 2\sqrt 2 \ and \ y_1 = \frac{5}{\sqrt 2}\)
⇒ \(\frac{{\sqrt 2x\;}}{{{8}}} + \frac{{\sqrt 2y\;}}{{{10}}} = 1\)
⇒ 5√2 x + 4√2 y = 40
So, the equation of the tangent to the given ellipse is: 5√2 x + 4√2 y = 40
Hence, option B is the correct answer.
