Application of Derivatives 1 Question 14

####14. On the ellipse $4 x^{2}+9 y^{2}=1$, the point at which the tangents are parallel to the line $8 x=9 y$, are $(1999,2 \mathrm{M})$

(a) $\frac{2}{5}, \frac{1}{5}$

(b) $-\frac{2}{5}, \frac{1}{5}$

(c) $-\frac{2}{5},-\frac{1}{5}$

(d) $\frac{2}{5},-\frac{1}{5}$

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Answer:

Correct Answer: 14. (8)

Solution:

  1. Given, $4 x^{2}+9 y^{2}=1$

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & 8 x+18 y \frac{d y}{d x}=0 \\ \Rightarrow \quad & \frac{d y}{d x}=-\frac{8 x}{18 y}=-\frac{4 x}{9 y} \end{aligned} $

The tangent at point $(h, k)$ will be parallel to $8 x=9 y$, then

$ -\frac{4 h}{9 k}=\frac{8}{9} $

$ \Rightarrow \quad h=-2 k $

Point $(h, k)$ also lies on the ellipse.

$ \therefore \quad 4 h^{2}+9 k^{2}=1 $

On putting value of $h$ in Eq. (ii), we get

$ \begin{array}{rlrl} & & 4(-2 k)^{2}+9 k^{2} & =1 \\ \Rightarrow & 16 k^{2}+9 k^{2} & =1 \\ \Rightarrow & & 25 k^{2} & =1 \\ \Rightarrow & k^{2} & =\frac{1}{25} \\ \Rightarrow & & k & = \pm \frac{1}{5} \end{array} $

Thus, the point, where the tangents are parallel to $8 x=9 y$ are $-\frac{2}{5}, \frac{1}{5}$ and $\frac{2}{5},-\frac{1}{5}$.

Therefore, options (b) and (d) are the answers.



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