Let P and Q be two points on the curves x2+y2=2 and (x2/8)+(y2/4)=1 respectively. Then the minimum value of the length PQ is

Question

Let P and Q be two points on the curves x2+y2=2 and (x2/8)+(y2/4)=1 respectively. Then the minimum value of the length PQ is (A) 1 (B) 2-√2 (C) 2√2 (D) √2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-thfdooy9jind.png" /> The shortest distance PQ occurs along the common normal; i.e., when P is (0 , √2) Q is (0,2). Hence (P Q) textminimum=2-√2

Q. Let $P$ and $Q$ be two points on the curves $x^{2} + y^{2} = 2$ and $\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1$ respectively. Then the minimum value of the length $PQ$ is

$1$

$2 - 2$

$22$

$2$

The shortest distance $PQ$ occurs along the common normal;
i.e., when $P$ is $(0, 2)$ & $Q$ is $(0, 2) .$
Hence $(PQ)_{minimum} = 2 - 2$

Q. Let P and Q be two points on the curves x2+y2=2 and 8x2​+4y2​=1 respectively. Then the minimum value of the length PQ is

1

2−2​

22​

2​

The shortest distance PQ occurs along the common normal; i.e., when P is (0,2​) & Q is (0,2). Hence (PQ)minimum​=2−2​

Q. Let $P$ and $Q$ be two points on the curves $x^{2} + y^{2} = 2$ and $\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1$ respectively. Then the minimum value of the length $PQ$ is

$1$

$2 - 2$

$22$

$2$

The shortest distance $PQ$ occurs along the common normal;
i.e., when $P$ is $(0, 2)$ & $Q$ is $(0, 2) .$
Hence $(PQ)_{minimum} = 2 - 2$