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Maths-

General

Easy

Question

The sum of distances of any point on the ellipse 3 x² + 4y² = 24 from its foci is

8 $\sqrt{2}$
4 $\sqrt{2}$
16 $\sqrt{2}$
none of these

hint Hint:

The sum of the distances to any point on the ellipse (x,y) from the two foci (c,0) and (-c,0) is a constant. That constant will be 2a. If we let d₁ and d₂ bet the distances from the foci to the point, then d₁ + d₂ = 2a.

The correct answer is: 4 $square root of 2$

Given : $3 x squared space plus space 4 y squared space equals space 24$
Dividing both sides by 24, we get
$rightwards double arrow fraction numerator 3 x squared over denominator 24 end fraction space plus space fraction numerator 4 y squared over denominator 24 end fraction space equals space 1 rightwards double arrow space x squared over 8 space plus space y squared over 6 space equals space 1 C o m p a r i n g space f r o m space g e n e r a l space f o r m space x squared over a squared space plus space y squared over b squared space equals 1 rightwards double arrow a space equals space square root of 8 space space a n d space b space equals space square root of 6$
The sum of the distances to any point on the ellipse (x,y) from the two foci (c,0) and (-c,0) is a constant. That constant will be 2a. If we let d1 and d2 bet the distances from the foci to the point, then d1 + d2 = 2a.
$rightwards double arrow$ d1 + d2 = 2a
$rightwards double arrow$ d1 + d2 = 2 $square root of 8$
$rightwards double arrow$ d1 + d2 = 4 $square root of 2$
Thus, the sum of distances of any point on the ellipse 3 x2 + 4y2 = 24 from its foci is 4 $square root of 2$ .

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