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>> Normal to Parabola-Part1
Normal to a Parabola
Normal at the point (x1, y1)
The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1). Since the slope of the tangent = 2a/y1, slope of the normal is –y1/2a. Also it passes through (x1, y1).
Hence, its equation is y – y1 = -y1/2a (x – x1). … (i)
Normal in terms of m
In equation (i), put -y1/2a = m so that y1 = –2a and x1 = (y12)/4a = am2, then the equation becomes y = mx – 2am – am3
where m is a parameter. Equation (ii) is the normal at the point (am2, –2am) of theparabola.
If this normal passes through a point (h, k), then k = mh – 2am – am3.
For a given parabola and a given point (h, k), this cubic in m has three roots say m1, m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes are m1, m2, m3. For this 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 the normals drawn from a point (h, k) to a given parabola y2 = 4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we can get the locus of (h, k).
Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the normals from a given point is zero. These points are called Co-Normal Points.
Find the locus of the point of intersection of two normals to a parabolawhich are at right angles to one another.
The equation of the normal to the parabola y2 = 4ax is
y = mx – 2am – am3.
It passes through the point (h, k) if
k = mh – 2am – am3 => am3 + m(2a – h) + k = 0. … (1)
Let the roots of the above equation be m1, m2and m3. Let the perpendicularnormals correspond to the values of m1 and m2 so that m1 m2 = –1.
From equation (1), m1 m2 m3 = -k/a. Since m1 m2 = –1, m3 = k/a.
Since m3 is a root of (1), we have a (k/a)3+k/a (2a – h) + k = 0. ⇒ k2 + a(2a – h) + a2 = 0
⇒ k2 = a(h – 3a).
Hence the locus of (h, k) is y2 = a(x – 3a).
Normal at the point t
The normal, being perpendicular to the tangent at (at, 2at) is given by y = –tx + 2at + at3.
If normal at the point at1 meets the parabola again at the point at2, then at2 = – at1 – 2/at1 .
Point of intersection of the normals to the parabola y2 = 4ax (at12, 2at1) and (at2t2, 2at2) is (2a+a(t12+t22+t1t2), a t1t2(t1+t2)).
Prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus.
If the normal to the parabola y2 = 4ax at P(at1t2, 2at1) meets it again at the point t2, then we have t2 = – t1 – 2/t1 .
If the abscissa and the ordinates of P be equal, then at12 = 2at1
⇒ t1 = 2 (rejecting t1 = 0) t2 = – 2 – 1 = – 3
The co-ordinates of P and Q are therefore (4a, 4a) and (9a, – 6a) respectively.
The focus is the point S (a, 0).
Slope of PS = ¾ and slope of QS = – ¾.
⇒ ∠PSQ = right angle. Hence the result.