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```        If a normal is drawn at a variable point P of the ellipse (X^2/a^2)+(y^2/b^2)=1,find the maximum distance of the normal from the centre of the ellipse.
```
7 years ago

510 Points
```										general point on ellipse is (x,y)= (acos@,bsin@)
x=acos@                              &                    y =bsin@
dx/d@=-asin@                     &                    dy/d@=bcos@
dy/dx=-bcot@/a
slope of tangent=dy/dx=-(bcot@)/a
slope of normal=-1/dy/dx =atan@/b
now equation of normal at (acos@,bsin@) is
yb-axtan@+a2sin@-b2sin@ =0
distance of this line from center of ellipse(0,0) is
d=  (a2-b2)sin@/(1+tan2@)1/2
=(a2-b2)sin@/sec@=(a2-b2)sin2@ /2
maximum value of sin2@ is 1
so maximum distance is dmax= (a2-b2)/2
```
7 years ago
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