Statement-I : The sum of the slopes of the tangents drawn from (5,4) to $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1 i s \frac{40}{9}$
Statement-II :The product of the slopes in the above situation is $\frac{4}{9}$
Which of the above statements is true

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neither I nor II

both I & II

only II

only I

answer is C.

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Detailed Solution

(I)

$G i v e n e l l i p s e \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1 w h e r e a^{2} = 16, b^{2} = 12 a n d t h e p o i n t = (5, 4) N o w m_{1} + m_{2} = \frac{2 x_{1} y_{1}}{{x_{1}}^{2} - a^{2}} = \frac{2 \times 5 \times 4}{25 - 16} = \frac{40}{9}$

(II)

$m_{1} m_{2} = \frac{{y_{1}}^{2} - b^{2}}{{x_{1}}^{2} - a^{2}} = \frac{16 - 12}{25 - 16} = \frac{4}{9}$

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