SATHEE: Application of Derivatives 1 Question 14

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 $(1999, 2 M)$

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

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

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

Show Answer

Answer:

Correct Answer: 14. (8)

Solution:

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

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

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

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

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

$\Rightarrow h = - 2 k$

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

$∴ 4 h^{2} + 9 k^{2} = 1$

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

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

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.

Application of Derivatives 1 Question 14

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 1 Question 14

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu