"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 :"

Question

Accepted Answer

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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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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 :

Solution

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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 :

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