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The tangents to the curve y=(x-2)^2-1 at its points of intersection with the line x-y=3,  intersect at the point

Option: 1

\begin{aligned} & \left(\frac{5}{2}, 1\right) \\ \end{aligned}


Option: 2

\left(-\frac{5}{2},-1\right) \\


Option: 3

\left(\frac{5}{2},-1\right) \\


Option: 4

\left(-\frac{5}{2}, 1\right)


Answers (1)

best_answer

Given equation of parabola is -

\begin{aligned} & y=(x-2)^2-1 \\ & \Rightarrow y=x^2-4 x+3 \end{aligned}........................(i)

Now, let \left(x_1, x_2\right) be the point of intersection of tangents of parabola (i) and line x-y=3, then

Equation of chord of contact of point \left(x_1, x_2\right) w.r.t. parabola (i) is 

\begin{aligned} & \Rightarrow \frac{1}{2}\left(y+y_1\right)=x x_1-2\left(x+x_1\right)+3 \\ & \Rightarrow\left(y+y_1\right)=2 x\left(x_1-2\right)-4 x_1+6 \\ & \Rightarrow 2 x\left(x_1-2\right)-y=4 x_1+y_1-6 \end{aligned}

this equation represent the line x-y=3 only, so on comparing, 

\begin{aligned} & \frac{2\left(x_1-2\right)}{1}=\frac{-1}{-1}=\frac{4 x_1+y_1-6}{3} \\ & \Rightarrow x_1=\frac{5}{2} \text { and } y_1=-1 \end{aligned}

So, the required point is \left(\frac{5}{2},-1\right)

Posted by

Deependra Verma

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