Question

Statement 1: Given the base BC of the triangle and the ratio radius of the ex-circles opposite to the angles B and C. Then locus of the vertex A is hyperbola.
Statement 2: $|S^{\prime }P-SP|=2a$, where S and S' are the two foci, $2a=$ length of the transverse axis and P be any point on the hyperbola.

• A. Both the statements are true and statement 2 is the correct explanation of statement 1.
• B. Both the statements 1 are true and statement 2 is not the correct explanation of statement 1.
• C. Statement 1 is true and statement 2 is false.
• D. Statement 1 is false and statement 2 is true.

Answer 1

The correct option is D Statement 1 is false and statement 2 is true.

Let P(x,y) be any point on the right branch of hyperbola and $S_1(-ae,0)$ and $S_2(ae,0)$ are foci
$P{S}_1^2=(x+ae{)}^2+y^2$
$=x^2+2axe+a^2{e}^2+(x^2-a^2)(e^2-1)$
$=x^2{e}^2+2axe+a^2$
$\Rightarrow P{S}_1^2=(ex+a{)}^2$
$\Rightarrow P{S}_1=ex+a$ .....(1)
$P{S}_2^2=(x-ae{)}^2+y^2$
$=x^2-2axe+a^2{e}^2+(x^2-a^2)(e^2-1)$
$=x^2{e}^2-2axe+a^2$
$\Rightarrow P{S}_1^2=(ex-a{)}^2$ (Since, P is on right branch of hyperbola , $ex>a$)
$\Rightarrow P{S}_2=ex-a$ ....(2)
Subtracting (2) from (1), we get
$P{S}_1-P{S}_2=2a$
If we consider P(x,y) on the left branch of hyperbola , then $ex<-a$
$P{S}_2-P{S}_1=2a$
So, $|P{S}_1-P{S}_2|=2a$
So, statement 2 is true.