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AP EAMCET · Maths · Straight Lines

If \(A=(a, 0)\) and \(B=(-a, 0)\), then the locus of a point \(\mathrm{P}\) such that \(\mathrm{PA}^2-\mathrm{PB}^2=a^2\) is.

  1. A a circle
  2. B an ellipse
  3. C a hyperbola
  4. D a straight line
Verified Solution

Answer & Solution

Correct Answer

(D) a straight line

Step-by-step Solution

Detailed explanation

Let \\(P=(x, y)\\). \\(PA^2 - PB^2 = a^2\\) \\((x-a)^2 + y^2 - ((x+a)^2 + y^2) = a^2\\) \\((x^2 - 2ax + a^2) - (x^2 + 2ax + a^2) = a^2\\) \\(-4ax = a^2\\) \\(x = -\frac{a}{4}\\) A straight line.
From AP EAMCET
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