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Let $\mathrm{ P(a, b)}$ be a point on the parabola $\mathrm{ y^{2}=8 x}$ such that the tangent at $\mathrm{ P}$ passes through the centre of the circle $\mathrm{ x^{2}+y^{2}-10 x-14 y+65=0.}$ Let $\mathrm{A }$ be the product of all possible values of $\mathrm{a }$ and $\mathrm{B }$ be the product of all possible values of $\mathrm{b. }$ Then the value of $\mathrm{ A+B }$ is equal to
Option: 1
$0$

Option: 2
$25$

Option: 3
$40$

Option: 4
$65$

Answers (1)

best_answer

$\mathrm{P(a,b)}$ is point on $\mathrm{y^{2}= 8x}$ , such that tangent at $\mathrm{P}$ pass through centre of

$\mathrm{\begin{aligned} x^{2}+y^{2}-10 x-14 y &+65 =0 \end{aligned}}$ i.e $(5,7)$

Tangent at $\mathrm{P\left(a t^{2}, 2 a t\right) \text { is ty }=x+a t^{2}}$

$\mathrm{A= 2 \& \text { it pass through }(5,7)} \\$

$\mathrm{7 t=5+2 t^{2}} \\$

$\mathrm{\Rightarrow t=1, t=\frac{5}{2}} \\$
$\mathrm{\therefore P\left(a t^{2}, 2 a t\right) \Rightarrow(2,4) \text { when } t=1} \\$ $\mathrm{\&\left(\frac{25}{2}, 10\right) \text { when } t=\frac{5}{2}} \\$

$\mathrm{\therefore A=2 \times \frac{25}{2}=25} \\$

$\mathrm{ \therefore A+B=65}$

Hence correct option is 4

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