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If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is :

JEE · Math · previous-year question

  1. A.2x – y – 2 = 0correct
  2. B.3x – 2y = 0
  3. C.2x – 3y + 10 = 0
  4. D.x – 2y + 8 = 0

Answer

A. 2x – y – 2 = 0

Explanation

Formula for standard hyperbola : $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ It passes through (4, 6) $$ \therefore $$ $${{16} \over {{a^2}}} - {{36} \over {{b^2}}} = 1$$ ........(1) We know, $${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$ $$ \Rightarrow $$ 4 = $$1 + {{{b^2}} \over {{a^2}}}$$ [ as e = 2 ] $$ \Rightarrow $$ $${{b^2} = 3{a^2}}$$ ....(2) Putting $${{b^2} = 3{a^2}}$$ in equation (1) $${{16} \over {{a^2}}} - {{36} \over {3{a^2}}} = 1$$ $$ \Rightarrow $$ $${{a^2} = 4}$$ $$ \therefore $$ , $${{b^2} = 12}$$ So Equation of hyperbola is $${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 1$$ Equation of tangent to the hyperbola at (4, 6) is $${{4x} \over 4} - {{6y} \over {12}} = 1$$ $$ \Rightarrow $$ $$x - {y \over 2} = 1$$ $$ \Rightarrow $$ 2x – y – 2 = 0

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