If the angle between the tangents drawn to th | vedclass

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Question

(A) Let the point on the line $4x - y = 0$ be $P(h, k)$. Since $P$ lies on the line,$k = 4h$. So,$P = (h, 4h)$.
The locus of points from which the an

Vedclass · Accepted Answer

$\frac{14}{47}$

Vedclass · Answer

(A) Let the point on the line $4x - y = 0$ be $P(h, k)$. Since $P$ lies on the line,$k = 4h$. So,$P = (h, 4h)$.
The locus of points from which the angle between the tangents to the parabola $y^2 = 4ax$ is $\alpha$ is given by the equation $(y^2 - 4ax) 	an^2 \alpha = (x + a)^2$.
Here,$a = 1$ and $\alpha = \frac{\pi}{3}$,so $	an^2 \alpha = (\sqrt{3})^2 = 3$.
The equation becomes $3(y^2 - 4x) = (x + 1)^2$.
Substituting $P(h, 4h)$ into this equation:
$3((4h)^2 - 4h) = (h + 1)^2$
$3(16h^2 - 4h) = h^2 + 2h + 1$
$48h^2 - 12h = h^2 + 2h + 1$
$47h^2 - 14h - 1 = 0$.
The sum of the abscissae $h$ is the sum of the roots of this quadratic equation,which is $-\frac{b}{a} = -(\frac{-14}{47}) = \frac{14}{47}$.
Wait,re-evaluating the question constraints: The options provided do not match $\frac{14}{47}$. Let's re-check the calculation. The locus of the intersection of tangents at angle $	heta$ is $(y^2 - 4ax) 	an^2 	heta = (x + a)^2$. With $a=1, 	heta = 60^\circ$,$3(y^2 - 4x) = (x+1)^2$. For $y=4x$,$3(16x^2 - 4x) = x^2 + 2x + 1 \implies 48x^2 - 12x = x^2 + 2x + 1 \implies 47x^2 - 14x - 1 = 0$. The sum is $14/47$. Given the options,there might be a typo in the question's line equation or angle. Assuming the question intended $x-y=0$ or similar,but based on provided text,the solution is $14/47$.

If the angle between the tangents drawn to the parabola $y^2 = 4x$ from a point on the line $4x - y = 0$ is $\frac{\pi}{3}$,then the sum of the abscissae of all such points is

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