Question

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

• A. $x+y-4=0$
• B. $x-y+4=0$
• C. $x-y-1=0$
• D. $x-y+1=0$

Answer 1

The correct option is D $x-y+1=0$

For the parabola $y^2=4x$, differentiating both sides w.r.t x, we get

$2y\frac{dy}{dx}=4$ or $\frac{dy}{dx}=\frac{2}{y}$

Since the tangent is inclined at $\frac{\pi }{4}$ in positive direction of x axis, $\frac{2}{y_1}=\tan \left(\frac{\pi }{4}\right)=1$

$\Rightarrow y_1=2$

Correspondingly, $x_1=1$

Thus, we need a line having slope 1 and passing through $(1,2)$

Which is $y-x-1=0$ or $x-y+1=0$