The tangents to the curve y = (x - 2)^2 - 1 a | vedclass

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Question

(D) The given curve is $y = (x - 2)^2 - 1$,which can be written as $(x - 2)^2 = y + 1$.Let the point of intersection of the tangents be $P(x_1, y_1)$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given curve is $y = (x - 2)^2 - 1$,which can be written as $(x - 2)^2 = y + 1$.Let the point of intersection of the tangents be $P(x_1, y_1)$.The chord of contact of the parabola $(x - 2)^2 = y + 1$ from the point $P(x_1, y_1)$ is given by the equation $T = 0$:$(x - 2)(x_1 - 2) = \frac{1}{2}(y + y_1 + 2)$$2(x - 2)(x_1 - 2) = y + y_1 + 2$$2x(x_1 - 2) - 4(x_1 - 2) = y + y_1 + 2$$2(x_1 - 2)x - y - (4x_1 - 8 + y_1 + 2) = 0$$2(x_1 - 2)x - y - (4x_1 + y_1 - 6) = 0 \quad ......(i)$We are given that the chord of contact is the line $x - y - 3 = 0 \quad ......(ii)$Comparing the coefficients of equations $(i)$ and $(ii)$:$\frac{2(x_1 - 2)}{1} = \frac{-1}{-1} = \frac{-(4x_1 + y_1 - 6)}{-3}$From $\frac{2(x_1 - 2)}{1} = 1$,we get $2x_1 - 4 = 1$ $\Rightarrow 2x_1 = 5$ $\Rightarrow x_1 = \frac{5}{2}$.From $\frac{-(4x_1 + y_1 - 6)}{-3} = 1$,we get $4x_1 + y_1 - 6 = 3 \Rightarrow 4x_1 + y_1 = 9$.Substituting $x_1 = \frac{5}{2}$ into the equation: $4(\frac{5}{2}) + y_1 = 9$ $\Rightarrow 10 + y_1 = 9$ $\Rightarrow y_1 = -1$.Thus,the point of intersection is $\left( \frac{5}{2}, -1 \right)$.

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:

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