Normal to the parabola:
Normal at Point (x1, y1 ):
The equation of the tangent at point (x1, y1) is yy1 = 2a(x + x1). As the slope of
tangent = 2a/y1 , slope of normal is -y1/ 2a . Also it passes through (x1, y1).
Therefore its equation is y - y1 = - y1/2a(x - x1) . . . . . (i)
Normal in Terms of m:
In the equation (i), put -y1/2a so that y1 = -2am and x1 = y12, then equation becomes y = mx - 2am - am3 . . . (ii)
where m is parameter. Equation (ii) is the normal at point (am2, -2am) of the parabola.
Notes:
- If this normal passes through the point (h, k), then k = mh - 2am - am3.
For the given parabola and a given point (h, k) , this cubic in m is having 3 roots say m1, m2, m3 that is from (h, k) 3 normals can be drawn to parabola whose slopes are m1, m2, m3 . For the cubic, we have
m1+ m2 + m3 = 0
m1 m2 +m2 m3 +m3 m1 = (2a-h) /a
m1 m2 m3 = - k/a
If we have an extra condition about normals drawn from the point (h, k) to the given parabola y2 =4ax then by eliminating m1, m2, m3 from these 4 relations between m1, m2, m3, we can get locus of (h, k).
- As the sum of the roots is equal to zero, the sum of ordinates of the feet of normals from a given point is zero.
Normal at the Point 't':
Equation of normal to y2 = 4ax at point (x1, y1) is y - y1 = - y1/2a(x - x1)
If (x1, y1) ≡ (at2, 2at)
Equation of normal becomes
y - 2at = -2at/2a(x - at2) => y = -tx + 2at + at3.
Notes:
- If normal at point t1 meets parabola again at the point t2, then t2 = -t1 -2/t1.
- Point of intersection of normals to parabola y2 = 4 ax at (at12, 2at1) and
(at22, 2at2) is (2a + a(t12 + t22 + t1t2),- at1t2(t1+ t2)) .
Example: Find out the locus of the foot of perpendicular drawn from the vertex on a tangent to parabola y2 = 4ax.
Solution: Any tangent of slope m to parabola y2 = 4ax is
y = mx + a/m . . . (1)
The perpendicular from vertex on (1) is
x + my = 0 . . . . (2)
By eliminating m between (1) and (2) we get the required locus as xy2 + x3 + ay2 = 0
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