Question

Find the point on the curve ax² + 2bxy + ay² = c, 0 < a < b < c, whose distance from the origin is minimum.

Answer 1

Let the point P be (rcosθ, rsinθ)

Given curve is ax² + 2bxy + ay² = c ...(i)

(i) reduces to ar²cos²θ + r²bsin2θ + ar²sin²θ = c

⇒  r² = c/(a cos²θ + bsin2θ + asin²θ)

⇒  r² =  c/(a + bsin2θ) 

⇒ r² will be minimum when (a + b sin2θ) is maximum

sin2θ = 1 ⇒ θ = π/4 or 5π/4

Therefore, r² = c/(a + b)

Thus, the points are

= (rcosθ, rsinθ)

= (rcos(π/4), rsin(π/4)), (rcos(5π/4), rsin (5π/4)) 

=  (r/√2 , r/√2), ( – r/√2 , – r/√2)