Question:
The normal at the point $(at_1^2 ,\, 2at_1)$ on the parabola meets the parabola again in the point $(at_1^2 ,\, 2at_2)$ , then
The normal at the point $(at_1^2 ,\, 2at_1)$ on the parabola meets the parabola again in the point $(at_1^2 ,\, 2at_2)$ , then
Updated On: Jun 17, 2022
- $ t_{2} = -t_{1} +\frac{2}{t_{1}}$
- $ t_{2} = -t_{1} -\frac{2}{t_{1}}$
- $ t_{2} = t_{1} -\frac{2}{t_{1}}$
- $ t_{2} = t_{1} +\frac{2}{t_{1}}$
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The Correct Option is B
Solution and Explanation
Equation of the normal at point $(at_1^2 ,\, 2at_1)$ on parabola is
$y = -t_{1}x + 2at_{1} + at_{1}^{3}$
It also passes through $\left(at_{1}^{2} ,\, 2at_{2}\right)$
So, $2at_{2} = -t_{1}\left(at^{2}_{2}\right)+2at_{1}+at^{3}_{1}$
$\Rightarrow 2t_{2}-2t_{1} = -t_{1}\left(t^{2}_{2}-t^{2}_{1}\right)$
$\Rightarrow t_{1}+t_{2} = \frac{-2}{t_{1}}$
$\Rightarrow t_{2} = -t_{1} -\frac{2}{t_{1}}$
$y = -t_{1}x + 2at_{1} + at_{1}^{3}$
It also passes through $\left(at_{1}^{2} ,\, 2at_{2}\right)$
So, $2at_{2} = -t_{1}\left(at^{2}_{2}\right)+2at_{1}+at^{3}_{1}$
$\Rightarrow 2t_{2}-2t_{1} = -t_{1}\left(t^{2}_{2}-t^{2}_{1}\right)$
$\Rightarrow t_{1}+t_{2} = \frac{-2}{t_{1}}$
$\Rightarrow t_{2} = -t_{1} -\frac{2}{t_{1}}$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.