P₁, P₂ . . . . ., P_i, . . . . . Pn are the points on the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ and Q₁, Q₂ , ....Q_i, . . . Q_nare the corresponding points on the auxiliary circle of the ellipse. If the line joining centre C to Q_i meets the normal at P_i w.r.t. the given ellipse at K_i and $\sum_{i = 1}^{n} ({CK}_{i}) = 175$ then the value of n/5 is

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answer is 5.

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Detailed Solution

$G i v e n e l l i p s e \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 w h e r e a^{2} = 16, b^{2} = 9$

Equation of normal at $P_{i} is \frac{4 x}{\cos θ_{i}} - \frac{3 y}{\sin θ_{i}} = 7$ …….(1)
Equation of the line joining C and Q_i is
$y = \frac{\sin θ_{i}}{\cos θ_{i}} . x$ .......(2) From (1) and (2)
$\begin{array}{l} \Rightarrow x = 7 \cos θ_{i} and y = 7 \sin θ_{i} \\ ∴ K_{i} = (7 \cos θ_{i}, 7 \sin θ_{i}) and C (0, 0) \end{array}$
(centre of the ellipse) Now . C.K _i= 7
$\begin{array}{l} \Rightarrow \sum_{i = 1}^{n} {CK}_{i} = 175 \\ \Rightarrow 7 n = 175 \Rightarrow n = 25 \\ \Rightarrow \frac{n}{5} = 5 \end{array}$