Question

PA and PB are the tangents drawn to y = 4ax from point P. These tangents meet y axis at the points A and B, respectively. If the area of ΔPAB is 2 sq. units, then locus of P is

• A. (y + 4ax)x = 16
• B. (y– 4ax)x = 8
• C. (y– 4ax)x = 16
• D. None of these

Answer 1

Answer: (C)

(y– 4ax)x = 16

Answer 2

The given question has a likely typo in the equation of the parabola. Assuming the parabola is $y^2 = 4ax$, similar to the provided example, we proceed as follows.

Let the point P be $(x_0, y_0)$.
Let the tangents from P to the parabola $y^2 = 4ax$ touch the parabola at $T_1(x_1, y_1)$ and $T_2(x_2, y_2)$.
The equation of the tangent at a point $(x', y')$ on the parabola $y^2 = 4ax$ is $yy' = 2a(x + x')$.

The tangent PA from P$(x_0, y_0)$ touches the parabola at $T_1(x_1, y_1)$. The equation of this tangent is $yy_1 = 2a(x + x_1)$. Since it passes through P$(x_0, y_0)$, we have $y_0y_1 = 2a(x_0 + x_1)$.
This tangent meets the y-axis at point A. Setting $x=0$ in the tangent equation, we get $yy_1 = 2ax_1$. So, the y-coordinate of A is $y_A = \frac{2ax_1}{y_1}$. Since $T_1(x_1, y_1)$ is on the parabola, $y_1^2 = 4ax_1$, which implies $x_1 = \frac{y_1^2}{4a}$. Substituting this into the expression for $y_A$, we get $y_A = \frac{2a(y_1^2/4a)}{y_1} = \frac{y_1^2/2}{y_1} = \frac{y_1}{2}$.
So, the point A is $(0, y_1/2)$.

Similarly, the tangent PB from P$(x_0, y_0)$ touches the parabola at $T_2(x_2, y_2)$. The equation of this tangent is $yy_2 = 2a(x + x_2)$. It passes through P$(x_0, y_0)$, so $y_0y_2 = 2a(x_0 + x_2)$.
This tangent meets the y-axis at point B. Setting $x=0$, we get $yy_2 = 2ax_2$, so $y_B = \frac{2ax_2}{y_2}$. Since $y_2^2 = 4ax_2$, we have $x_2 = \frac{y_2^2}{4a}$. Substituting this, we get $y_B = \frac{2a(y_2^2/4a)}{y_2} = \frac{y_2}{2}$.
So, the point B is $(0, y_2/2)$.

The points forming the triangle are P$(x_0, y_0)$, A$(0, y_1/2)$, and B$(0, y_2/2)$.
The base AB of the triangle lies on the y-axis. The length of the base is $|y_B - y_A| = |\frac{y_2}{2} - \frac{y_1}{2}| = \frac{1}{2}|y_2 - y_1|$.
The height of the triangle from P to the y-axis is the absolute value of the x-coordinate of P, which is $|x_0|$.
The area of $\triangle PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2}|y_2 - y_1| \times |x_0| = \frac{1}{4}|x_0(y_2 - y_1)|$.
We are given that the area is 2 sq. units.
$\frac{1}{4}|x_0(y_2 - y_1)| = 2 \implies |x_0(y_2 - y_1)| = 8$.

The points of tangency $T_1(x_1, y_1)$ and $T_2(x_2, y_2)$ are the intersections of the parabola $y^2 = 4ax$ and the chord of contact from P$(x_0, y_0)$, whose equation is $yy_0 = 2a(x + x_0)$.
From the chord of contact equation, $x = \frac{yy_0 - 2ax_0}{2a}$. Substitute this into the parabola equation:
$y^2 = 4a \left( \frac{yy_0 - 2ax_0}{2a} \right) = 2(yy_0 - 2ax_0) = 2yy_0 - 4ax_0$.
$y^2 - 2y_0y + 4ax_0 = 0$.
The roots of this quadratic equation in $y$ are $y_1$ and $y_2$, the y-coordinates of the points of tangency $T_1$ and $T_2$.
By Vieta's formulas:
$y_1 + y_2 = 2y_0$
$y_1 y_2 = 4ax_0$

We need $|y_2 - y_1|$. We can find $(y_2 - y_1)^2$:
$(y_2 - y_1)^2 = (y_1 + y_2)^2 - 4y_1y_2 = (2y_0)^2 - 4(4ax_0) = 4y_0^2 - 16ax_0 = 4(y_0^2 - 4ax_0)$.
So, $|y_2 - y_1| = \sqrt{4(y_0^2 - 4ax_0)} = 2\sqrt{|y_0^2 - 4ax_0|}$.
For real tangents to exist from P$(x_0, y_0)$, P must lie outside or on the parabola, which means $y_0^2 - 4ax_0 \ge 0$. Thus, $|y_0^2 - 4ax_0| = y_0^2 - 4ax_0$.
$|y_2 - y_1| = 2\sqrt{y_0^2 - 4ax_0}$.

Substitute this into the area equation $|x_0(y_2 - y_1)| = 8$:
$|x_0| (2\sqrt{y_0^2 - 4ax_0}) = 8$.
$|x_0| \sqrt{y_0^2 - 4ax_0} = 4$.
Squaring both sides:
$x_0^2 (y_0^2 - 4ax_0) = 16$.

The locus of P$(x_0, y_0)$ is obtained by replacing $(x_0, y_0)$ with $(x, y)$:
$x^2 (y^2 - 4ax) = 16$.

This equation can be written as $x^2 y^2 - 4ax^3 = 16$.
Let's check the given options, assuming the question meant $y^2 = 4ax$.
a) $(y + 4ax)x = 16 \implies xy + 4ax^2 = 16$
b) $(y – 4ax)x = 8 \implies xy - 4ax^2 = 8$
c) $(y – 4ax)x = 16 \implies xy - 4ax^2 = 16$
d) None of these

None of the options match the derived locus $x^2(y^2 - 4ax) = 16$. Let's re-examine the similar question's solution. The similar question involves $y^2 = 4x$ (so $a=1$) and the locus is $a^2 (y^2 - 4x) x^2 = 16$. Substituting $a=1$, we get $1^2 (y^2 - 4x) x^2 = 16$, which is $x^2(y^2 - 4x) = 16$. This matches our derived locus for the general parabola $y^2 = 4ax$ with $a=1$.
So, if the parabola in the given question was $y^2 = 4ax$, the locus is $x^2(y^2 - 4ax) = 16$. This is not among the options.

Let's consider the possibility that the parabola is $x^2 = 4ay$. We derived the locus $x^4(x^2 - 4ay) = 16a^2$ in the thought process. This is also not among the options.

Let's consider the possibility that the parabola is $y = 4ax$. As discussed, this is a line, and tangents in the usual sense do not exist from an external point.

Given the context of JEE/NEET and conic sections, it is highly probable that the equation of the parabola is intended to be $y^2=4ax$ or $x^2=4ay$. Since the similar question uses $y^2=4x$, it is most likely that the intended parabola is $y^2=4ax$. If the parabola is $y^2=4ax$, the locus is $x^2(y^2 - 4ax) = 16$. This is not in the options.

Given the high probability of a typo in the parabola equation in the question and the mismatch with the options, and the consistency of our derivation with the similar question's result (if the parabola was $y^2=4ax$), it's highly likely that the question or options have errors.

However, if we are forced to choose from the given options, it suggests that either the problem is based on a different parabola or the options are incorrect, or there is a fundamental misunderstanding of the problem statement due to typos.