Question

# The equation of the tangent to the parabola $$y^2=4x$$ inclined at an angle of $$\dfrac{\pi}{4}$$ to the +ve direction of $$x-axis$$ is.

A
x+y4=0
B
xy+4=0
C
xy1=0
D
xy+1=0

Solution

## The correct option is D $$x-y+1=0$$For the parabola $$y^2 = 4x$$, differentiating both sides w.r.t x, we get $$2y \cfrac{dy}{dx} = 4$$ or $$\cfrac{dy}{dx} = \cfrac{2}{y}$$Since the tangent is inclined at $$\cfrac{\pi}{4}$$ in positive direction of x axis, $$\cfrac{2}{y_1} = \tan \left ( \cfrac{\pi}{4} \right ) = 1$$$$\Rightarrow y_1 = 2$$Correspondingly, $$x_1 = 1$$Thus, we need a line having slope 1 and passing through $$(1,2)$$Which is $$y - x - 1 = 0$$ or $$x - y + 1 = 0$$Mathematics

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