Foci of an ellipse are at S(1, 7), S1(1, –3). The point P is on the ellipse such that SP = 7, S1P = 5. Then the equation of the ellipse is

( x − 1 ) 2 11 + ( y − 2 ) 2 36 = 1

Foci of an ellipse are at S(1, 7), S1(1, –3). The point P is on the ellipse such that SP = 7, S1P = 5. Then the equation of the ellipse is

( x − 1 ) 2 11 + ( y − 2 ) 2 36 = 1

( x − 1 ) 2 11 + ( y + 2 ) 2 36 = 1

( x − 1 ) 2 11 + ( y − 2 ) 2 36 = 1

( x − 1 ) 2 36 + ( y − 2 ) 2 11 = 1

( x − 1 ) 2 11 + ( y − 2 ) 2 6 = 1

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Foci of an ellipse are at S(1, 7), S¹(1, –3). The point P is on the ellipse such that SP = 7, S¹P = 5. Then the equation of the ellipse is

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$\frac{(x - 1)^{2}}{11} + \frac{(y + 2)^{2}}{36} = 1$

$\frac{(x - 1)^{2}}{11} + \frac{(y - 2)^{2}}{36} = 1$

$\frac{(x - 1)^{2}}{36} + \frac{(y - 2)^{2}}{11} = 1$

$\frac{(x - 1)^{2}}{11} + \frac{(y - 2)^{2}}{6} = 1$

answer is B.

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Detailed Solution

$F r o m t h e g i v e n d a t a a x i s i s p a r a l l e l t o y - a x i s t h e n SP + S^{'} P = 2 b \Rightarrow b = 6, 2 b e = 10 \Rightarrow e = \frac{5}{6} \Rightarrow a^{2} = b^{2} (1 - e^{2}) \Rightarrow a^{2} = 11$
Ellipse $\frac{(x - 1)^{2}}{11} + \frac{(y - 2)^{2}}{36} = 1$