Multiple choice

If chord contact of the tangents drawn from the point $\left( \alpha ,\beta \right) $ to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1,$ touches the circle ${ x }^{ 2 }+{ y }^{ 2 }={ c }^{ 2 },$ THEN THE LOCUS OF THE POINT $\left( \alpha ,\beta \right) $

  1. $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =\dfrac { 1 }{ { c }^{ 2 } } $
  2. $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =\dfrac { 1 }{ { c }^{ 4 } } $
  3. $\dfrac { { x }^{ 2 } }{ { a }^{ 4 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 4 } } =\dfrac { 1 }{ { c }^{ 2 } } $

  4. none of these
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C Correct answer
Explanation

The chord of contact of tangents from (alpha, beta) to the ellipse x^2/a^2 + y^2/b^2 = 1 is (alpha*x)/a^2 + (beta*y)/b^2 = 1. If this line touches the circle x^2 + y^2 = c^2, the perpendicular distance from the origin to the line must equal the radius c. Thus, 1 / sqrt(alpha^2/a^4 + beta^2/b^4) = c, leading to alpha^2/a^4 + beta^2/b^4 = 1/c^2.

AI explanation

The equation of the chord of contact from the point (alpha, beta) to the ellipse x squared over a squared plus y squared over b squared equals 1 is x alpha over a squared plus y beta over b squared equals 1. Since this line touches the circle x squared + y squared equals c squared, the perpendicular distance from the origin (0,0) to the line must equal the radius c. Equating the distance formula gives 1 divided by the square root of (alpha squared over a to the fourth power plus beta squared over b to the fourth power) equals c. Squaring and rearranging this relationship yields the locus x squared over a to the fourth power plus y squared over b to the fourth power equals 1 over c squared.

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