Porisma

Worked solution

Ellipse: find the farthest point

Sketch the curve 4x2+y2+2y−3=04x^2+y^2+2y-3=0 and write down the equation(s) of its line(s) of symmetry and the maximum distance of the point (0,5)(0,5) to the curve.

[2012/CJC/Promo/2]

CJC · Discussion · Question 2

Answer

Symmetry lines: x=0x=0, y=−1y=-1. Maximum distance: 88 units, attained at (0,−3)(0,-3).

Sketch and symmetry

4x2+(y+1)2−1−3=04x^2+(y+1)^2-1-3=0 gives x2+(y+1)2/4=1x^2+(y+1)^2/4=1. The centre is (0,−1)(0,-1) with endpoints (±1,−1),(0,1),(0,−3)(\pm1,-1),(0,1),(0,-3). The symmetry lines are x=0x=0 and y=−1y=-1.

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Maximum distance

The farthest point is (0,−3)(0,-3), giving 5−(−3)=85-(-3)=8 units as in the source. To verify that no off-axis point is farther, set x=cos⁡tx=\cos t, y=−1+2sin⁡ty=-1+2\sin t. The squared distance is cos⁡2t+(−6+2sin⁡t)2=37−24s+3s2\cos^2t+(-6+2\sin t)^2=37-24s+3s^2, where s=sin⁡t∈[−1,1]s=\sin t\in[-1,1]. Its derivative with respect to ss is −24+6s<0-24+6s<0 throughout this interval, so its maximum is at s=−1s=-1, with squared distance 6464.

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Ellipse with centre (0,−1), endpoints (±1,−1),(0,1),(0,−3). The segment from external point (0,5) to bottom endpoint (0,−3) has length 8, the maximum distance.
Ellipse with centre (0,−1), endpoints (±1,−1),(0,1),(0,−3). The segment from external point (0,5) to bottom endpoint (0,−3) has length 8, the maximum distance.
Ellipse with centre (0,−1), endpoints (±1,−1),(0,1),(0,−3). The segment from external point (0,5) to bottom endpoint (0,−3) has length 8, the maximum distance.