CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
xy+4=0
loader
C
xy1=0
loader
D
xy+1=0
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image