Question

The number of normals that can be drawn through the point (2,0) to the parabola $y^2 = 7x$ is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For a parabola $y^2=4ax$, the number of distinct normals that can be drawn from a point $(h,k)$ is determined by the number of real roots of the cubic equation $k = mh - 2am - am^3$. A simpler condition is that if $h \leq 2a$, there is always exactly one real normal.

Answer 1

The Correct Option is B

The equation of the normal to the parabola $y^2=4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.
For the given parabola $y^2 = 7x$, we have $4a=7$, so $a=7/4$.
The equation of the normal becomes $y = mx - 2(\frac{7}{4})m - (\frac{7}{4})m^3 = mx - \frac{7}{2}m - \frac{7}{4}m^3$.
This normal passes through the point $(h,k) = (2,0)$. Substitute $x=2, y=0$:
$0 = m(2) - \frac{7}{2}m - \frac{7}{4}m^3$.
$0 = 2m - \frac{7}{2}m - \frac{7}{4}m^3$.
$0 = -\frac{3}{2}m - \frac{7}{4}m^3$.
Multiply by $-4$: $0 = 6m + 7m^3$.
Factor out $m$: $m(7m^2+6) = 0$.
This gives two possibilities for the slope $m$.
Case 1: $m=0$. This is a real value for the slope.
Case 2: $7m^2+6=0 \implies 7m^2 = -6 \implies m^2 = -6/7$. This gives no real solutions for $m$.
Since there is only one real value for the slope ($m=0$), only one normal can be drawn from the point (2,0) to the parabola.
Alternatively, for a point $(h,k)$, the number of normals depends on the condition $27ak^2<4(h-2a)^3$. Here $a=7/4, h=2, k=0$. $27(7/4)(0)^2 = 0$. $4(h-2a)^3 = 4(2-2(7/4))^3 = 4(2-7/2)^3 = 4(-3/2)^3 = 4(-27/8) = -27/2$. The condition $0<-27/2$ is false. If $h>2a$, which is $2>2(7/4) = 3.5$ (false), there is one normal. Let's check the condition again. The condition for three distinct normals from $(h,k)$ is $h>2a$ and $27ak^2<4(h-2a)^3$. Here $h=2$ and $2a = 7/2 = 3.5$. Since $h<2a$, there is only one normal.