Home
Class 12
MATHS
Maximum length of chord of the ellipse (...

Maximum length of chord of the ellipse `(x^(2))/(8)+(y^(2))/(4)=1`, such that eccentric angles of its extremities differ by `(pi)/(2)` is :

Text Solution

Verified by Experts

The correct Answer is:
`4.00`
Let the ends be `(2 sqrt(2) cos alpha, 2 sin alpha)` and `(-2sqrt(2)sin alpha, 2cos alpha)` then
`l = sqrt(8(cos alpha + sin alpha)^(2)+4(sin alpha - cos alpha)^(2))`
`= sqrt(12+4sin 2alpha)`
`l_("max")=sqrt(16)=4`
Promotional Banner

Similar Questions

Explore conceptually related problems

The points on the ellipse (x^(2))/(25)+(y^(2))/(9)=1 Whose eccentric angles differ by a right angle are

Find the sum of squares of two distances of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 whose extremities have eccentric angles differ by (pi)/(2) and sow that is a constant.

Find the locus of the foot of perpendicular from the centre of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 on the chord joining the points whose eccentric angles differ by (pi)/(2)

P and Q are two points on the ellipse (x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1 whose eccentric angles are differ by 90^(@) , then

Find the length of the chord x-2y-2=0 of the ellipse 4x^(2)+16y^(2)=64

Locus of mid-point of the focal chord of ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 with eccentricity e is

Find the condition so that the line px+qy=r intersects the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 in points whose eccentric angles differ by (pi)/(4) .

The locus of the point of intersection of tangents to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 at the points whose eccentric angles differ by pi//2 , is