The point on the curve y = 6x − x2 at which the tangent to the curve is inclined at $\pi$/4 to the line x + y= 0 is __________ .

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(3, 9) Let (x1, y1) be the point where the given curve intersect the given line at the given angle. \text { Since, the point lie on the curve } . \text { Hence }, y_1 = 6 x_1 - {x_1}^2 \text { Now,} y = 6x - x^2 \Rightarrow dy/dx = 6 - 2x \Rightarrow m_1 = 6 - 2 x_1 \text { and } x + y = 0 \Rightarrow 1 + dy/dx = 0 \Rightarrow dy/dx = - 1 \Rightarrow m_2 = - 1 \text { It is given that the angle between them is }$\pi$/4. \therefore \tan $\theta$ = \left| m_1 - m_2/1 + m_1 m_2 \right| \Rightarrow \tan $\pi$/4 = \left| 6 - 2 x_1 + $\frac{1}{1}$ - 6 + 2 x_1 \right| \Rightarrow 1 = \left| 7 - 2 x_1/2 x_1 - 5 \right| \Rightarrow 7 - 2 x_1/2 x_1 - 5 = $\pm$ 1 \Rightarrow 7 - 2 x_1/2 x_1 - 5 = 1 \ or7 - 2 x_1/2 x_1 - 5=-1 \Rightarrow 7 - 2 x_1 = 2 x_1 - 5 \ or \ 7 - 2 x_1 = - 2 x_1 + 5 \Rightarrow 4 x_1 = 12 \ or \ 2 = 0 (\text {It is not true }.) \Rightarrow x_1 = 3 \text { and } y_1 = 6 x_1 - {x_1}^2 = 18 - 9 = 9 \therefore\left( x_1 , y_1 \right)=\left( 3, 9 \right)

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