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

(a) (−3, −27)
(b) (3, 9)
(c) (7/2, 35/4)
(d) (0, 0)

Answer 1

(b) (3, 9)

Let (x₁, y₁) be the point where the given curve intersect the given line at the given angle.
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=6x_1-{x_1}^2 \\ \mathrm{Now},y=6x-x^2 \\ \Rightarrow \frac{dy}{dx}=6-2x \\ \Rightarrow m_1=6-2x_1 \\ \mathrm{and} \\ x+y=0 \\ \Rightarrow 1+\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-1 \\ \Rightarrow m_2=-1 \\ \text{It is given that the angle between them is}\frac{\mathrm{\pi }}{4}\text{.} \\ \therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right| \\ \Rightarrow \tan \frac{\mathrm{\pi }}{4}=\left|\frac{6-2x_1+1}{1-6+2x_1}\right| \\ \Rightarrow 1=\left|\frac{7-2x_1}{2x_1-5}\right| \\ \Rightarrow \frac{7-2x_1}{2x_1-5}=\pm 1 \\ \Rightarrow \frac{7-2x_1}{2x_1-5}=1\text{or}\frac{7-2x_1}{2x_1-5}\text{=-1} \\ \Rightarrow 7-2x_1=2x_1-5\text{or}7-2x_1=-2x_1+5 \\ \Rightarrow 4x_1=12\text{or 2 = 0 (It is not true.)} \\ \Rightarrow x_1=3 \\ \mathrm{and} \\ {y}_1=6x_1-{x_1}^2=18-9=9 \\ \text{∴}\left(x_1,y_1\right)\text{=}\left(3,9\right) \end{gathered}$$