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The tangents to the curve \mathrm{ y=(x-2)^2-1} at its points of intersection with the line x-y=3, intersect at the point \mathrm{\left(\frac{m}{n},-1\right). }Then m-n is

Option: 1

-7


Option: 2

-3


Option: 3

7


Option: 4

3


Answers (1)

best_answer

The given curve is Y
\mathrm{y=(x-2)^2-1} and the given line is x – y = 3.

Now, put x – 2 = X and y + 1 = Y. Equation of curve becomes

\mathrm{Y=X^2}                   .....[i]

and equation of line becomes Y = X ...(ii)
So, the points of intersection of (i) and (ii) are O(0, 0) and
A(1, 1).
Now, tangent at O(0, 0) to the curve \mathrm{Y=X^2} is X-axis, i.e.,
Y = 0.
Equation of tangent at A(1, 1) to the curve \mathrm{Y=X^2} is
Y + 1 = 2X.

\mathrm{\text { So, point of intersection of tangents is }\left(\frac{1}{2}, 0\right) \text {. }}

\mathrm{\therefore x-2=\frac{1}{2} \text { and } y+1=0 \Rightarrow x=\frac{5}{2} \text { and } y=-1}

Hence, the point of intersection of the given curve

\mathrm{y=(x-2)^2-1 \text { and the line } x-y=3 \text { is }\left(\frac{5}{2},-1\right) \text {. }}

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jitender.kumar

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