Question

The tangent and normal at the point $P(4,4)$ to the parabola , $y^2=4x$ intersects the x-axis at the points Q and R, respectively. Then the circumcenter of the $\Delta PQR$ is

• A. $(2,0)$
• B. $(2,1)$
• C. $(1,0)$
• D. $(1,2)$

Answer 1

The correct option is C $(1,0)$

Given curve is,

$y2=4x⟹2y\frac{dy}{dx}=4⟹\frac{dy}{dx}=\frac{2}{y}$

Slope of tangent at P(4,4)=$\frac{2}{4}=\frac{1}{2}$

Slope of normal=$\frac{-2}{1}$

Equation of tangent

$y-4=\frac{1}{2}(x-4)⟹2y-8=x-4⟹x-2y+4=0-\left(1\right)$

Equation of normal

$y-4=-2(x-4)⟹y-4=-2x+8⟹2x+y-12=0‘-\left(2\right)$

Since eqs.(1) and (2) intersect the x-axis i.e. $y=0$

we get $x=-4$ for tangent and $x=6$ for normal

So the coordinates of triangle are (4,4), (-4,0) and (6,0)

Since tangent and normal are perpendicular to each other

So circumcenter lies iin middle of the hypotenuse

circumcenter=$(\frac{-4+6}{2},\frac{0+0}{2})=(1,0)$