Question

Assertion :If $A$ and $B$ are two fixed points and $P$ a variable point, the locus of $P$ such that $|PA-PB|=k$ is a hyperbola for all $k$. Reason: If $S_1$ and $S_2$ are foci of a hyperbola and P be a point on the hyperbola then $|P{S}_1-P{S}_2|<S_1{S}_2$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

A Hyperbola is defined as the locus of all those points such that the difference of the distances of those points from two fixed points (also known as foci) is a constant and it's value is always less than the distance between two fixed points. ( i.e. foci )

Hence if $A$ and $B$ are the two fixed points and $P$ is a variable point, then the locus of $P$ such that $|PA-PB|=k,$ is a hyperbola. But $k$ cann't belong to all real numbers.

From the definition of the hyperbola, $k<\text{distance between two given fixed points}\left(AB\right)$

Hence Assertion is Incorrect.

Also If $S_1$ and $S_2$ are foci of hyperbola and $P$ be a point on the hyperbola then by the definition of hyperbola we know that,

$|P{S}_1-P{S}_2|=\text{length of major axis}=2a$

For any hyperbola the distance between the foci $S_1{S}_2=2ae,$ where $e$ is the eccentricity of the hyperbola and $e>1$

$\because 2a<2ae,$

$\therefore |P{S}_1-P{S}_2|<S_1{S}_2$

Hence the reason is correct too. But you can see that Assertion is incorrect.

$\Rightarrow$ Hence Correct answer is $D$.