SATHEE: Application of Derivatives 4 Question 42

Application of Derivatives 4 Question 42

####44. Find a point on the curve $x^{2} + 2 y^{2} = 6$ whose distance from the line $x + y = 7$ , is minimum.

$(2003, 2 M)$

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Answer:

Correct Answer: 44. $2 h k$

Solution:

Let us take a point $P (\sqrt{6} \cos θ, \sqrt{3} \sin θ)$ on $\frac{x^{2}}{6} + \frac{y^{2}}{3} = 1$ .

Now, to minimise the distance from $P$ to given straight line $x + y = 7$ , shortest distance exists along the common normal.

Slope of normal at $P = \frac{a^{2} / x_{1}}{b^{2} / y_{1}} = \frac{\sqrt{6} \sec θ}{\sqrt{6} cosec θ} = \sqrt{2} \tan θ = 1$

So, $\cos θ = \sqrt{\frac{2}{3}}$ and $\sin θ = \frac{1}{\sqrt{3}}$

Hence, required point is $P (2, 1)$ .

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Application of Derivatives 4 Question 42

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