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.
$(2003,2 \mathrm{M})$
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Answer:
Correct Answer: 44. $2 h k$
Solution:
- Let us take a point $P(\sqrt{6} \cos \theta, \sqrt{3} \sin \theta)$ 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 $P=\frac{a^{2} / x_{1}}{b^{2} / y_{1}}=\frac{\sqrt{6} \sec \theta}{\sqrt{6} \operatorname{cosec} \theta}=\sqrt{2} \tan \theta=1$
So, $\quad \cos \theta=\sqrt{\frac{2}{3}}$ and $\sin \theta=\frac{1}{\sqrt{3}}$
Hence, required point is $P(2,1)$.
