Home
Class 12
MATHS
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))`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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

If the tangent at the point P(theta) to the ellipse 16 x^2+11 y^2=256 is also a tangent to the circle x^2+y^2-2x=15 , then theta= (a) (2pi)/3 (b) (4pi)/3 (c) (5pi)/3 (d) pi/3

If the tangent at the point (4 cos theta, (16)/(sqrt(11)) sin theta) to the ellipse 16x^(2) + 11y^(2) = 256 is also a tangent to the circle x^(2) + y^(2) - 2x - 15 = 0 , then the value of theta , is

if the tangent at the point (4 cos phi , (16)/(sqrt(11) )sin phi ) to the ellipse 16x^(2)+11y^(2) =256 Is also a tangent to the circle x^(2) +y^(2)-2x=15, then the value of phi is

Find the number of common tangents to the circle x^2 +y^2=4 and x^2+y^2−6x−8y−24=0

The number of common tangents of the circles x^2+y^2−2x−1=0 and x^2+y^2−2y−7=0

Lengths of common tangents of the circles x^(2)+y^(2)=6x,x^(2)+y^(2)+2x=0 are

Find the points on the curve 2a^2y=x^3-3a x^2 where the tangent is parallel to x-axis.

Show that if the length of the tangent from a point P to the circle x^2 + y^2 = a^2 be four times the length of the tangent from it to the circle (x-a)^2 + y^2 = a^2 , then P lies on the circle 15x^2 + 15y^2 - 32ax + a^2=0 .

If the tangent to the ellipse x^2 +4y^2=16 at the point 0 sanormal to the circle x^2 +y^2-8x-4y=0 then theta is equal to