Question:

The normal chord at a point 't' on the parabola $ y^2 = 4ax$ subtends a right angle at the vertex. Then $t^2$ =

Updated On: Jul 7, 2022

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The Correct Option is B

Solution and Explanation

Normal at $'t'$ for parabola $ y^2 = 4ax$ is $tx + y = 2ax + ax^3 \quad...(1)$ Combined equation of the lines joining the vertex i.e., origin to the pts. of intersection of the parabola and $(1)$ is $y^{2} = 4ax\left(\frac{tx+y}{2at+at^{3}}\right) $ $ \Rightarrow \left(2t+t^{3}\right)y^{2}=4x\left(y+tx\right) $ Since $\left(1\right)$ makes a right angle at the vertex $\therefore 2t+ t^{3} - 4t = 0$ $\Rightarrow t^{2} = 2$

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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.