Question

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

विकल्प

• (−3, −27)

• (3, 9)

• (7/2, 35/4)

• (0, 0)

Answer 1

`(3, 9)`

 

Let (x₁, y₁) be the point where the given curve intersect the given line at the given angle.

\[\text { Since, the point lie on the curve } . \]

\[\text { Hence }, y_1 = 6 x_1 - {x_1}^2 \]

\[\text { Now,} y = 6x - x^2 \]

\[ \Rightarrow \frac{dy}{dx} = 6 - 2x\]

\[ \Rightarrow m_1 = 6 - 2 x_1 \]

\[\text { and }\]

\[x + y = 0\]

\[ \Rightarrow 1 + \frac{dy}{dx} = 0 \]

\[ \Rightarrow \frac{dy}{dx} = - 1\]

\[ \Rightarrow m_2 = - 1\]

\[\text { It is given that the angle between them is }\frac{\pi}{4}.\]

\[ \therefore \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\]

\[ \Rightarrow \tan \frac{\pi}{4} = \left| \frac{6 - 2 x_1 + 1}{1 - 6 + 2 x_1} \right|\]

\[ \Rightarrow 1 = \left| \frac{7 - 2 x_1}{2 x_1 - 5} \right|\]

\[ \Rightarrow \frac{7 - 2 x_1}{2 x_1 - 5} = \pm 1\]

\[ \Rightarrow \frac{7 - 2 x_1}{2 x_1 - 5} = 1 \ or\frac{7 - 2 x_1}{2 x_1 - 5}=-1\]

\[ \Rightarrow 7 - 2 x_1 = 2 x_1 - 5 \ or \ 7 - 2 x_1 = - 2 x_1 + 5\]

\[ \Rightarrow 4 x_1 = 12 \ or \ 2 = 0 (\text {It is not true }.)\]

\[ \Rightarrow x_1 = 3\]

\[\text { and }\]

\[ y_1 = 6 x_1 - {x_1}^2 = 18 - 9 = 9\]

\[\therefore\left( x_1 , y_1 \right)=\left( 3, 9 \right)\]