Question

Find the locus of the point P from which tangents are drawn to the parabola $y^2=4ax$ having slopes $m_1$ and $m_2$ such that - (i) $m_1^2 + m_2^2 = 10$ (ii) $\theta_1 - \theta_2 = 30^\circ$ (constant) where $\theta_1$ and $\theta_2$ are the inclinations of the tangents from positive x-axis.

• A. The locus is a discrete set of points given by $29x^2 - 8ax - a^2 = 0$ and $y^2 = 10x^2 + 2ax$.
• B. The locus is $y^2 = 10x^2 + 2ax$.
• C. The locus is $3y^2 = x^2 + 14ax + a^2$.
• D. The locus is the union of $y^2 = 10x^2 + 2ax$ and $3y^2 = x^2 + 14ax + a^2$.

Answer 1

Answer: (A)

The locus is a discrete set of points given by $29x^2 - 8ax - a^2 = 0$ and $y^2 = 10x^2 + 2ax$.

Answer 2

The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$. If this tangent passes through a point $P(x, y)$, then $m^2x - my + a = 0$. This is a quadratic equation in $m$, whose roots are the slopes $m_1$ and $m_2$ of the tangents from $P(x, y)$ to the parabola.

From Vieta's formulas: Sum of slopes: $m_1 + m_2 = \frac{y}{x}$ Product of slopes: $m_1 m_2 = \frac{a}{x}$

Let $\theta_1$ and $\theta_2$ be the inclinations of the tangents with the positive x-axis. Then $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$.

Condition (i): $m_1^2 + m_2^2 = 10$ $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1m_2$ $(\frac{y}{x})^2 - 2(\frac{a}{x}) = 10$ $\frac{y^2}{x^2} - \frac{2a}{x} = 10$ Multiplying by $x^2$: $y^2 - 2ax = 10x^2$ $y^2 = 10x^2 + 2ax$ (Locus 1)

Condition (ii): $\theta_1 - \theta_2 = 30^\circ$ $\tan(\theta_1 - \theta_2) = \tan(30^\circ) = \frac{1}{\sqrt{3}}$ $\frac{m_1 - m_2}{1 + m_1 m_2} = \frac{1}{\sqrt{3}}$ We know $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = (\frac{y}{x})^2 - 4(\frac{a}{x}) = \frac{y^2 - 4ax}{x^2}$. So, $m_1 - m_2 = \pm \frac{\sqrt{y^2 - 4ax}}{x}$ (for real tangents, $y^2 - 4ax \ge 0$). Substituting into the tangent subtraction formula: $\frac{\pm \frac{\sqrt{y^2 - 4ax}}{x}}{1 + \frac{a}{x}} = \frac{1}{\sqrt{3}}$ $\frac{\pm \sqrt{y^2 - 4ax}}{x+a} = \frac{1}{\sqrt{3}}$ Squaring both sides: $\frac{y^2 - 4ax}{(x+a)^2} = \frac{1}{3}$ $3(y^2 - 4ax) = (x+a)^2$ $3y^2 - 12ax = x^2 + 2ax + a^2$ $3y^2 = x^2 + 14ax + a^2$ (Locus 2)

The question asks for the locus of P where both conditions are satisfied. This means we need to find the intersection of Locus 1 and Locus 2. Substitute $y^2$ from Locus 1 into Locus 2: $3(10x^2 + 2ax) = x^2 + 14ax + a^2$ $30x^2 + 6ax = x^2 + 14ax + a^2$ $29x^2 - 8ax - a^2 = 0$

This quadratic equation gives the possible x-coordinates of the point P. For each valid x, the y-coordinate is determined by Locus 1: $y^2 = 10x^2 + 2ax$. Therefore, the locus of P is a discrete set of points satisfying both equations. The equation $29x^2 - 8ax - a^2 = 0$ defines the x-coordinates, and $y^2 = 10x^2 + 2ax$ defines the corresponding y-coordinates. This is the most accurate representation of the locus under both conditions.