If the point P on the curve, 4x² + 5y² = 20 is
farthest from the point Q(0, -4), then PQ² is equal to:

Solution

Maths sir · 4 steps

Write PQ² in terms of y alone, then remember y is trapped between -2 and 2 on this ellipse.

1. Remove x. From 4x² + 5y² = 20 you get x² = 5 - 5/4y². On this ellipse y can only take values from -2 to 2.

   PQ² = x² + (y+4)² = 5 - 5/4y² + y² + 8y + 16 = 21 + 8y - y²/4

2. Check the slope. The derivative is 8 - y/2, which is zero only at y = 16. That lies outside [-2, 2], so on the ellipse the derivative is always positive and PQ² keeps increasing with y.
3. Take the endpoint. The largest value is at y = 2, which gives x = 0 and P = (0, 2). Then PQ² = 21 + 16 - 1 = 36, and PQ = 6.
4. So. The farthest point is the top of the ellipse, and PQ² = 36, option (a).

The idea to keep. When a distance on a conic reduces to a function of one variable, always check that variable's allowed range. If the critical point falls outside it, the maximum sits at an endpoint.

Answer from the NTA final answer key. Solution by PrepWiser.