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Find the point on the ellipse 16 x^2+11 ...

Find the point on the ellipse `16 x^2+11 y^2=256` where the common tangent to ti and the circle `x^2+y^2-2x=15` toch.

Text Solution

Verified by Experts

The given ellipse is
`(x^(2))/(16)+(y^(2))/((256//11))=1`
Tangent to it at point `P( 4 cos theta, (16 sqrt(11)) sin theta) "is" (x)/(4) cos theta+(y)/((16//sqrt(11)))sin theta=1`
It also touches the circle `(x-1)^(2)+y^(2)=16`. Therefore, `(|(1)/(4)cos theta-1|)/(sqrt((cos^(2)theta)/(16)+(11 sin^(2) theta)/256))=4`
`or (|cos theta-4|)/(sqrt(16 cos^(2) theta+11 sin ^(2) theta))=1`
`or cos^(2)theta-8costheta+16=11+5cos^(2)theta`
or `4 cos^(2)theta+8cos theta-5=0`
or `(2 cos theta-1)(2 cos theta+5)=0`
or `cos theta =(1)/(2) or theta=+-(pi)/(3)`
Therefore, point are `(2,+-sqrt(3)//sqrt(11))`.
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