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 $P(a \cos \theta, 2 \sin \theta)$ be a point on the ellipse
$$ 4 x^{2}+a^{2} y^{2}=4 a^{2} \text {, i.e. } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1 $$
Let $A(0,-2)$ be the given point.
Then,
$$ \begin{aligned} (A P)^{2} & =a^{2} \cos ^{2} \theta+4(1+\sin \theta)^{2} \\ \Rightarrow \quad \frac{d}{d \theta}(A P)^{2} & =-a^{2} \sin 2 \theta+8(1+\sin \theta) \cdot \cos \theta \\ \Rightarrow \quad \frac{d}{d \theta}(A P)^{2} & =\left[\left(8-2 a^{2}\right) \sin \theta+8\right] \cos \theta \end{aligned} $$
For maximum or minimum, we put $\frac{d}{d \theta}(A P)^{2}=0$
$$ \begin{aligned} & \Rightarrow \quad\left[\left(8-2 a^{2}\right) \sin \theta+8\right] \cos \theta=0 \\ & \Rightarrow \quad \cos \theta=0 \quad \text { or } \quad \sin \theta=\frac{4}{a^{2}-4} \end{aligned} $$
$\left[\because 4<a^{2}<8 \Rightarrow \frac{4}{a^{2}-4}>1 \Rightarrow \sin \theta>1\right.$, which is
Now, $\frac{d^{2}}{d \theta^{2}}(A P)^{2}=-{\left(8-2 a^{2}\right) \sin \theta+8 } \sin \theta$
$$ +\left(8-2 a^{2}\right) \cdot \cos ^{2} \theta $$
For $\theta=\frac{\pi}{2}$, we have $\frac{d^{2}}{d \theta^{2}}(A P)^{2}=-\left(16-2 a^{2}\right)<0$
Thus, $A P^{2}$ i.e. $A P$ is maximum when $\theta=\frac{\pi}{2}$. 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 $a \cos \frac{\pi}{2}, 2 \sin \frac{\pi}{2}=(0,2)$
