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Question: The length of the portion of the normal at \(\left( {1,1} \right)\) to \({x^2} = y\) intercepted bet...

The length of the portion of the normal at (1,1)\left( {1,1} \right) to x2=y{x^2} = y intercepted between the axes is
A. 532\dfrac{{5\sqrt 3 }}{2}
B. 352\dfrac{{3\sqrt 5 }}{2}
C. 534\dfrac{{5\sqrt 3 }}{4}
D. 354\dfrac{{3\sqrt 5 }}{4}

Explanation

Solution

We will first find the slope of the normal to the curve x2=y{x^2} = y. Then, form the equation of the line normal to the given curve and pass through (1,1)\left( {1,1} \right). Next, find the coordinates of xx and yy intercepts. Hence, apply the distance formula to find the distance between intercepts.

Complete step-by-step answer:
We are given that the equation of the curve is x2=y{x^2} = y.
We will find the coordinates of normal at (1,1)\left( {1,1} \right)
We will find dydx\dfrac{{dy}}{{dx}} at point (1,1)\left( {1,1} \right) to find the normal.
Now, we know that d(xn)dx=nxn1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}
Then,
d(x2)dx=2x\dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2x
That is , y=2xy' = 2x
Hence the slope of the normal at (1,1)\left( {1,1} \right) is 12\dfrac{{ - 1}}{2}
Then, the equation of line passing through (1,1)\left( {1,1} \right)
(y1)=12(x1) 2y2=x+1 2y+x=3  \left( {y - 1} \right) = - \dfrac{1}{2}\left( {x - 1} \right) \\\ \Rightarrow 2y - 2 = - x + 1 \\\ \Rightarrow 2y + x = 3 \\\
We will now find xx intercept by putting yy equals to 0 in the above equation.
x=3x = 3
Similarly, we will now find yy intercept by putting xx equals to 0 in the above equation.
2y=3 y=32  2y = 3 \\\ y = \dfrac{3}{2} \\\
Hence, the coordinates of xx axis are (3,0)\left( {3,0} \right) and coordinates of yy axis are (0,32)\left( {0,\dfrac{3}{2}} \right).
We have to find the distance between (3,0)\left( {3,0} \right) and (0,32)\left( {0,\dfrac{3}{2}} \right)
Also, if (p,q)\left( {p,q} \right) and (m,n)\left( {m,n} \right) are two points, then the distance between them is given by (pm)2+(qn)2\sqrt {{{\left( {p - m} \right)}^2} + {{\left( {q - n} \right)}^2}}
Therefore, the distance between (3,0)\left( {3,0} \right) and (0,32)\left( {0,\dfrac{3}{2}} \right) is
(30)2+(032)2=9+94=452=352\sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {0 - \dfrac{3}{2}} \right)}^2}} = \sqrt {9 + \dfrac{9}{4}} = \dfrac{{\sqrt {45} }}{2} = \dfrac{{3\sqrt 5 }}{2}
Hence, option B is correct.

Note: If a line is passing through (x1,y1)\left( {{x_1},{y_1}} \right) and the slope of the line is mm, then the equation of the line is given as yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right). Also, the product of slope of perpendicular lines is equal to 1 - 1. xxintercept is the distance from the origin to the point where the line cuts the xx axis.