Multiple choice

If chord of contact of the tangent drawn from the point $(\alpha,\, \beta)$ to the ellipse $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, 1$ touches the circle $x^2 + y^2 = k^2$, then find the locus of the point $(\alpha,\, \beta)$.

  1. $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, \displaystyle \frac {1}{k^2}$
  2. $\displaystyle \frac {x^2}{a^4}\, +\, \displaystyle \frac {y^2}{b^4}\, =\, \displaystyle \frac {1}{k^2}$
  3. $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, \displaystyle k^2$
  4. $\displaystyle \frac {x^2}{a^4}\, +\, \displaystyle \frac {y^2}{b^4}\, =\, \displaystyle k^2$
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B Correct answer
Explanation

Chord of contact from (alpha, beta) to x^2/a^2 + y^2/b^2 = 1 is (alpha*x)/a^2 + (beta*y)/b^2 = 1. This line touches x^2 + y^2 = k^2. The condition for a line lx + my = 1 to touch x^2 + y^2 = k^2 is k^2(l^2 + m^2) = 1. Here l = alpha/a^2 and m = beta/b^2. So k^2(alpha^2/a^4 + beta^2/b^4) = 1. Thus, alpha^2/a^4 + beta^2/b^4 = 1/k^2.

AI explanation

Using the standard equation of the chord of contact from an external point to an ellipse, the equation of the chord of contact is xx1/a^2 + yy1/b^2 = 1. For this line to touch the circle x^2 + y^2 = k^2, the perpendicular distance from the origin to the line must equal the radius. The perpendicular distance from the origin to the line is 1 / sqrt((x1/a^2)^2 + (y1/b^2)^2). Equating this distance to k gives 1 / sqrt((x1/a^2)^2 + (y1/b^2)^2) = k. Squaring both sides yields x1^2/a^4 + y1^2/b^4 = 1/k^2. Therefore, the required locus is x^2/a^4 + y^2/b^4 = 1/k^2.

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