Question

Intersection point of tangent at ends of latus rectum of parabola $y^2=4x$ is

• A. $(1,0)$
• B. $(-1,0)$
• C. $(0,1)$
• D. $(0,-1)$

Answer 1

Answer: (D)

$(0,-1)$

The given equation of the parabola is $y^2=4x$

$4ax=4xa=1$

Equation of latus rectum is $x=1$

Putting $x=1$ in equation of parabola

we get

$y^2=4y=\pm 2$

Hence end point of latus rectum are $(1,2),(1,-2)$

Now equation of parabola is

$y^2=4x$

Differentiating on both sides with respect tox

$2y\cfrac{dy}{dx}=4\\ \Rightarrow\cfrac{dy}{dx}

=m=\cfrac{4}{2y}$

At $y=2m=1$

At $y=-2am=-1$

The equation of tangent is given by

$(y-y_1)=m(x-x_1)$

At $(1,2)$

$\Rightarrow y-2=1(x-1)\Rightarrow y=x+1$

At $(1,-2)y+2=-1(x-1)y=-x-1$

Intersection point of two tangent will be

$-x-1=x+1\Rightarrow 2x=-2\Rightarrow x=-1$

Now, $y=-x-1\Rightarrow y=1-1\Rightarrow y=0$

Hence, intersection point $(-1,0)$ lies on direction $x=-1$