Given two fixed points $A$ and $B$ and $A B = 6$ , then the simplest form of the equation to the locus of $P$ such that $P A + P B = 8$ is:

\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1

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\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

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\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1

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\frac{x^{2}}{12} + \frac{y^{2}}{21} = 1

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Solution

The correct option is C $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$
$A B =$ distance between foci $= 2 a e = 6$
$P A + P B = 2 a = 8$
The locus of $P$ is an ellipse with
$e = \frac{3}{4}$
and $a = 4$
$b^{2} = a^{2} - a^{2} e^{2} = 16 - 9 = 7$
So, the equations becomes $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$

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A (4, 0)

B (- 4, 0)

P A + P B = 10

\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1

Equation of the hyperbola with foci (0,± 5) and e = $\frac{5}{3}$ is

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1

Equation of the hyperbola with length of the latusrectum $\frac{9}{2}$ and e = $\frac{5}{4}$ is

E_{1} : \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

E_{2} : \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1

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