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The line $4 x-7 y+10=0$ intersects the parabola $y^2=4 a x$ , at the points A & B . The co-ordinates of the point of intersection of the tangents drawn at the points A & B are :

Option: 1
$(53,83)$

Option: 2
$(23,76)$

Option: 3
$(63,82)$

Option: 4
$(48,26)$

Answers (1)

best_answer

Line $4 x-7 y+10=0$ intersects the parabola $y^2=4 a x$

$\begin{aligned} & \Rightarrow y^2=7 y-10 \\ & \text { or } y^2-7 y+10=0 \\ & \Rightarrow(y-5)(y-2)=0 \\ & \therefore y=5,2 \end{aligned}$

$Put \: \: y=5 \: \: in\: \: y^2=4 x\: \: we \: \: get\\$

$4 x=25 \text { Or } x=\frac{25}{4}$

Again put $y=2 \: \: in\: \: y^2=4 x \: \: we \: \: get\: \: 4 x=4$

$or \: \: x=1\\ \therefore \mathrm{A}(1,2) \: \: and\: \: \mathrm{B}\left(\frac{25}{4}, 5\right)$

Tangent of $y^2=4 x\: \: is\: \: T=0 \: \: is \: \: y y_1=2\left(x+x_1\right)$

$At \: \: A (1,2) \: \: we \: \: have \: \: 2 y=2(x+1)\\$

$or \: \: y=x+1 \: \: or\: \: x-y+1=0$ is the equation of the tangent at the point A

$At \: \: B \left(\frac{25}{4}, 5\right) we\: \: have \: \: 5 y=2\left(x+\frac{25}{4}\right)$

$or \: \: 20 y=8 x+50 \: \: or \: \: \: 4 x-10 y-25=0$ is the equation of the tangent at the point B

$\begin{aligned} & \Rightarrow 10 x+10=4 x+25 \\ & \Rightarrow 10 x-4 x=25-10 \end{aligned}$

$\Rightarrow 6 x=15\\$

$or\: \: x=53$

Substituting $x=53 \: \: in\: \: x-y+1=0$ we get

$\begin{aligned} & 53-y+1=0 \\ & \Rightarrow-y+83=0 \\ & \therefore y=83 \end{aligned}$

∴ tangent intersect at the point $(53,83)$

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Rakesh

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