Slope of the tangent to the parabola $y^2=x+2$ at a point in $1^{st}$ quadrant and lying on the lines y = x is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{4}$
$y^2=x+2$
differentiating wrt x
$2y\frac{dy}{dx}=1⇒\frac{dy}{dx}=\frac{1}{2y}$ (slope)
P(x, y) lies in 1st quadrant and lies on x = y as well as $y^2=x+2$
$⇒x^2=x+2⇒(x-2)(x+1)=0$
$⇒x=2,y=2$ → 1st quadrant
Slope = $\frac{1}{2y}=\frac{1}{4}$