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(B) Let $P(h, k)$ be the point from which two tangents are drawn to $y^2 = 4x$. Any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{

Vedclass · Accepted Answer

$9x = 2y^2$

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(B) Let $P(h, k)$ be the point from which two tangents are drawn to $y^2 = 4x$. Any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$. If it passes through $P(h, k)$,then $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$. Let $m_1$ and $m_2$ be the roots of this quadratic equation. From the properties of roots,we have $m_1 + m_2 = \frac{k}{h}$ and $m_1 m_2 = \frac{1}{h}$. Given that $m_1 = 2m_2$,we substitute this into the equations: $2m_2 + m_2 = 3m_2 = \frac{k}{h} \Rightarrow m_2 = \frac{k}{3h}$. $2m_2 	imes m_2 = 2m_2^2 = \frac{1}{h} \Rightarrow m_2^2 = \frac{1}{2h}$. Substituting $m_2$ from the first into the second: $2(\frac{k}{3h})^2 = \frac{1}{h}$ $\Rightarrow 2(\frac{k^2}{9h^2}) = \frac{1}{h}$ $\Rightarrow \frac{2k^2}{9h} = 1$ $\Rightarrow 2k^2 = 9h$. Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $2y^2 = 9x$.

If two tangents drawn from a point $P$ to the parabola $y^2 = 4x$ are such that the slope of one tangent is double the other,then $P$ lies on the curve:

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