Question
Question: If \[P\] is a point on the ellipse \[\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1\] whose foci ...
If P is a point on the ellipse 16x2+25y2=1 whose foci are S and S′ , then PS+PS′=8 .
Solution
Some of the properties of an ellipse that we need to know before getting into this problem: For an ellipse a2x2+b2y2=1 , the sum of the distance of a point on the ellipse from the two foci will be equal to the length of major axis that is 2b . On find the length of major axis we can compare and find whether the given statement is true or false.
Complete Step by step answer:
It is given that P is a point on the ellipse 16x2+25y2=1 . The foci of this ellipse are S and S′ .
We aim is to verify that, PS+PS′=8 which is nothing but the sum of the focal distance from a moving point P.
The given ellipse equation is 16x2+25y2=1 . We know that the general equation of an ellipse is a2x2+b2y2=1 . Thus, we get a2=16 and b2=25 .
Now let us find the values of a & b which we will be using later.
We have a2=16 . Taking square root, we get a=±4 .
Likewise, we have b2=25 . Taking square root, we get b=±5 .
We know that the sum of the distance of a point on the ellipse from the two foci will be equal to the length of major axis that is 2b so, let us find the value of 2b first.
We know that the value of b is ±5 . Since the measure of length cannot be negative, we take b=5 .
Thus 2b=2×5=10 .
Thus, the length of major axis is 10 . By the statement that the sum of the distance of a point on the ellipse from the two foci will be equal to the length of major axis that is 2b , we have to get PS+PS′=10 , but it is given that PS+PS′=8 .
When we compare this, we get to know that the given PS+PS′=8 is wrong.
Note:
This problem is given like a statement thus, we have to verify it. When we verified this problem, we came to know that the given result is wrong. In case the given result is wrong then, we have to find the correct result and re-write the statement.