SATHEE: Application of Derivatives 4 Question 52

Application of Derivatives 4 Question 52

####55. Find the point on the curve $4 x^{2} + a^{2} y^{2} = 4 a^{2}, 4 < a^{2} < 8$ that is farthest from the point $(0, - 2)$ .

(1987, 4M)

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Answer:

Correct Answer: 55. (9)

Solution:

Let $P (a \cos θ, 2 \sin θ)$ be a point on the ellipse

$4 x^{2} + a^{2} y^{2} = 4 a^{2}, i.e. \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1$

Let $A (0, - 2)$ be the given point.

Then,

$\begin{aligned} (A P)^{2} & = a^{2} \cos^{2} θ + 4 (1 + \sin θ)^{2} \\ \Rightarrow \frac{d}{d θ} (A P)^{2} & = - a^{2} \sin 2 θ + 8 (1 + \sin θ) \cdot \cos θ \\ \Rightarrow \frac{d}{d θ} (A P)^{2} & = [(8 - 2 a^{2}) \sin θ + 8] \cos θ \end{aligned}$

For maximum or minimum, we put $\frac{d}{d θ} (A P)^{2} = 0$

$\begin{aligned} \Rightarrow [(8 - 2 a^{2}) \sin θ + 8] \cos θ = 0 \\ \Rightarrow \cos θ = 0 or \sin θ = \frac{4}{a^{2} - 4} \end{aligned}$

$[∵ 4 < a^{2} < 8 \Rightarrow \frac{4}{a^{2} - 4} > 1 \Rightarrow \sin θ > 1$ , which is

Now, $Missing or unrecognized delimiter for \left$

$+ (8 - 2 a^{2}) \cdot \cos^{2} θ$

For $θ = \frac{π}{2}$ , we have $\frac{d^{2}}{d θ^{2}} (A P)^{2} = - (16 - 2 a^{2}) < 0$

Thus, $A P^{2}$ i.e. $A P$ is maximum when $θ = \frac{π}{2}$ . The point on the curve $4 x^{2} + a^{2} y^{2} = 4 a^{2}$ that is farthest from the point

$A (0, - 2)$ is $a \cos \frac{π}{2}, 2 \sin \frac{π}{2} = (0, 2)$

Application of Derivatives 4 Question 52

Answer:

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Application of Derivatives 4 Question 52

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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