1.

The maximum distance of the centre of the ellipse `x^2/(81)+y^2/(25)=1` from a normal to the ellipse is

Answer» let us take a point as `(9cos theta , 5sin theta)`
the tangent on the point will be ` (x*9cos theta)/81 + (y*5*sin theta)/25 = 1 `
`(x*Cos theta)/9 + (y*sin theta)/5 = 1`
`y= (-5*cos theta)/(9*sin theta)* x + 5/ sin theta`
so slope will be `(5Cos theta)/ (9 sin theta)`
as we know that, slope of tangent * slope of normal = -1
so, slope of normal is `(9*Sin theta)/(5*cos theta)`
for normal `y- sin theta = (9*sin theta)/(5* cos theta) * (x- 9*cos theta)`
`5*y*cos theta - 25*sin theta*cos theta = 9*x*sin theta - 81* sin theta`
`5*y*cosec theta - 25 = 9*x*sec theta - 81`
`9*x*sec theta - 5*y*cosec theta - 56=0 `
` now, for centre (0,0)`
`|(0-0-56)/sqrt(81*cosec^2 theta +25*cosec^2 theta)| = 56/sqrt(81*sec^2 theta + 25*cosec^2 theta) `
sloving for denominator
let `sqrt(81*sec^2 theta + 25*cosec^2 theta) = x`
for maximum value the condition must be true that `(dx)/(d theta) = 0`
So, differentiating we get, ` (81*sec^2 theta * tan theta - 25* 2*cosec^2 theta * cot theta)/ (2*sqrt(81*sec^2 theta + 25* cosec^2 theta)) = 0`
`81*sec^2 theta * tan theta = 25* cosec^2 theta * cot theta`
`(81*sin theta)/ cos ^ 3 theta = 25*cos theta / sin ^ 3 theta`
`Sin^4 theta/ cos^4 theta = 25/81`
`tan^2 theta = 5/9`
`cot^2 theta = 9/5`
now putting in main equation
`56/sqrt(81*(1+tan^2 theta) + 25*(1+ cot^2 theta)) `
`= 56/ sqrt(81*(1+5/9) + 25*(1+9/5))`
`= 56/sqrt(9*14+ 5*14)`
`= 56/ 14 = 4`


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