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Let $$P(a, b)$$ be a point on the parabola $$y^{2}=8 x$$ such that the tangent at $$P$$ passes through the centre of the circle $$x^{2}+y^{2}-10 x-14 y+65=0$$. Let $$A$$ be the product of all possible values of $$a$$ and $$B$$ be the product of all possible values of $$b$$. Then the value of $$A+B$$ is equal to :

JEE · Math · previous-year question

  1. A.0
  2. B.25
  3. C.40
  4. D.65correct

Answer

D. 65

Explanation

Centre of circle $${x^2} + {y^2} - 10x - 14y + 65 = 0$$ is at (5, 7). Let the equation of tangent to $${y^2} = 8x$$ is $$yt = x + 2{t^2}$$ which passes through (5, 7) $$7t = 5 + 2{t^2}$$ $$ \Rightarrow 2{t^2} - 7t + 5 = 0$$ $$t = 1,{5 \over 2}$$ $$A = 2 \times {1^2} \times 2 \times {\left( {{5 \over 2}} \right)^2} = 25$$ $$B = 2 \times 2 \times 1 \times 2 \times 2 \times {5 \over 2} = 40$$ $$A + B = 65$$

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