Question

If F1 = (3, 0), F2 = (–3, 0) and P is any point on the curve 16x2 +25y2 = 400, then PF1 + PF2 equal

Solution

Correct option is

10

The equation of the ellipse can be written as

               

Here       a2 = 25, b2 = 16 

But         b2 = a2(1 – e2)

⇒             16 = 25(1 – e2)  

So that foci of the ellipse are (±ae, 0) i.e. (±3, 0) or F1 and F2

By definition of the ellipse, since P is any point on the ellipse

                PF1 + PF2 = 2a = 2 × 5 = 10

SIMILAR QUESTIONS

Q1

Consider the two curves C1 : y2 = 4x ; C2 : x2 + y2 – 6x + 1, then 

Q2

The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is

Q3

The normal at an end of a latus rectum of the ellipse  passes through an end of the minor axis if

Q4

If an ellipse slides between two perpendicular straight line, then the locus of its center is   

Q5

If the tangent at a point on the ellipse  meets the auxillary circle in two points, the chords joining them subtends a right angle at the center; then the eccentricity of the ellipse is given by

Q6

The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse. x2 + 9y2 = 9, meets the auxillary circle at the point M, then the area of the triangle with vertices at A, M and the origin is

Q7

The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q, then locus of M intersects the latus rectums of the given ellipse at the points.

Q8

Let a and b be non-zero real numbers. Then the equation (ax2 + by2 + c)(x2 – 5xy + 6y2) = 0 represents

Q9

The pints of intersection of the two ellipse x2 + 2y2 – 6x – 12y + 23 = 0 and 4x2 + 2y2 – 20x – 12y + 35 = 0.

Q10

The tangent at any point P of the hyperbola  makes an intercept of length p between the point of contact and the transverse axis of the hyperbola, p1p2 are the lengths of the perpendiculars drawn from the foci on the normal at P, then