Question

A line passing through the point (21,30) and normal to the curve \[y = 2\sqrt x \] can have the slope:
A. 2
B. 3
C. -2
D. -5

Answer 1

Hint: The given curve is a parabola. Then find the equation of its normal using its slope. By using coordinates of the given point in the normal equation we will get an algebraic equation in m. Solving it we will get the required result.

Complete step-by-step answer:
The given curve is:
 \[y = 2\sqrt x \] ….. (1)
Taking square on both sides, we get
${y^2} = 4x$ …. (2)
which is a parabola along the x-axis.
As we know that the equation of the normal of the parabola ${y^2} = 4ax$, with slope m is :
$y = mx - 2am - a{m^3}$
Here a = 1. So above equation will be

$y = mx - 2m - {m^3}$ …..(3)
Since this normal line is passing through the point (21,30).
We put x=21 and y=30 in equation (3) to get equation for ‘m’, which will be
${m^3} - 19m + 30 = 0$….(4)
We have to solve the above equation (4) to find values of m.
We assume that one root is m=3. This value is satisfying the equation (4) as,
$
  {3^3} - 19 \times 3 + 30 = 0 \\
   \Rightarrow 0 = 0 \\
 $
Thus (m-3) is one root. We divide the equation (4) by the term (m-3) to get other factors as:
$\dfrac{{{m^3} - 19m + 30}}{{m - 3}} = {m^2} + 3m - 10$
Now, we will factorize ${m^2} + 3m - 10$ , as follows:
$
  {m^2} + (5 - 2)m - 10 \\
   \Rightarrow {m^2} + 5m - 2m - 10 \\
   \Rightarrow (m - 2)(m + 5) \\
 $
Therefore all three factors are (m-3) , (m-2) and (m+5) .
So, values of m will be, m = 2, 3, -5
Since x in the equation \[y = 2\sqrt x \] must be positive, hence y will also be positive. Therefore in the first quadrant its normal line will have slope angle more than ${90^0}$ and thus slope must be negative.
Therefore the valid value of slope m will be -5.

So, the correct answer is “Option D”.

Note: This problem is very interesting related to the equation of parabola and its normal line. Actually parabola will have different normal lines with different slopes depending on the portion of curve considered. Careful visualization of the curve will solve the problem in an easy way. Also, utilization of algebraic simplification rules are needed for the solution.