y=e^2 x+x^2 (d y)/(d x)=2 e^2 x+2 x at x=0, (d y)/(d x)=2 -(d x)/(d y)=-(1)/(2) eqn of normal

step 1 So in this question we have to find distance between origin and normal to the curve. Now we will

Y=e^2 x+x^2 (d y)/(d x)=2 e^2 x+2 x at x=0, (d y)/(d x)=2...

$y=e^{2 x}+x^{2}$
$\frac{d y}{d x}=2 e^{2 x}+2 x$
at $x=0, \quad \frac{d y}{d x}=2$
$-\frac{d x}{d y}=-\frac{1}{2}$
eqn of normal $\rightarrow y-1=-\frac{1}{2} x$
$\Rightarrow 2 y+x-2=0$
Distance from $(0,0)=\frac{2}{\sqrt{5}}$

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Key Concepts

Differentiation

Differentiation is the process of finding the derivative, which represents the rate of change of a function with respect to its variable. It involves applying rules such as the chain rule and power rule to compute derivatives of composite and polynomial functions.

Tangent and Normal Lines

The tangent line to a curve at a given point touches the curve exactly at that point and has a slope equal to the derivative of the function. The normal line at that point is perpendicular to the tangent and its slope is the negative reciprocal of the tangent’s slope.

Point-Slope Form of a Line

The point-slope form is a method used to write the equation of a line when a point on the line and its slope are known. It is expressed as y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope. This form is especially useful in deriving equations of tangents or normals.

Distance from a Point to a Line

The distance from a given point to a line can be calculated using the formula |ax0 + by0 + c| / ?(a² + b²) where the line is given by ax + by + c = 0 and the point by (x0, y0). This formula measures the shortest (perpendicular) distance from the point to the line.

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