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:
- 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.
