Question:

If the normal at point P $(a t_{1}^{2}, 2 a t_{1})$ and $Q$ $(a t_{2}^{2}, 2 a t_{2})$ on the parabola $y^{2} = 4 a x$ meet on the parabola, the $t_{1} t_{2}$ equals

JEE Main

Updated On: Oct 10, 2024

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Correct Answer: 2

Solution and Explanation

Explanation:
Given: Normals at

P

and

Q

at the points

(a t_{1}^{2}, 2 a t_{1})

and

(a t_{2}^{2}, 2 a t_{2})

respectively meet the parabola

y^{2} = 4 a x

We have to find the value of

t_{1} t_{2}

.
Let normal at

P

meets on the parabola

y^{2} = 4 a x

at

(a t^{2}, 2 a t) . .

then, by the property of normal to the parabola we have,

T = - t_{1} - \frac{2}{t_{1}}

.......(i)
Similarly, normal at Q meets on the parabola

y^{2} = 4 a x

at

(a t^{2}, 2 a t)

So,

T = - t_{2} - \frac{2}{t_{2}}

....(ii)
Using (i) and (ii), we get

\Rightarrow t_{1} - \frac{2}{t_{1}} = - t_{2} - \frac{2}{t_{2}}

(t_{2} - t_{1}) = 2 (\frac{1}{t_{1}} - \frac{1}{t_{2}})

\Rightarrow t_{1} t_{2} = 2 [∵ t_{1} \neq t_{2}]

Hence, the correct answer is 2.00.

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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).

Parabola

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.