Question

The locus of point of intersection of the two tangents to the parabola $y^2=4ax$ which intercept a given distance $4c$ on the tangent at the vertex is

• A. $y^2-4ax=\frac{c^2}{4}$
• B. $y^2-4ax=8c^2$
• C. $y^2-4ax=\frac{c^2}{2}$
• D. $y^2-4ax=16c^2$

Answer 1

Answer: (D)

$y^2-4ax=16c^2$

The vertex of the above parabola is the origin.

Hence the tangent to the above parabola at the origin is the y-axis.

Since the tangents make and intercept of 4c on the y axis, therefore, let the equation of the tangents be

$y=m_{1}x+4c$
$y=m_{2}x+4c$

Subtracting equation ii from i, we get

$(m_1-m_2)x=0$
$x=0$

Substituting x=0, we get the point of intersection as

$P=(0,4c)$

Hence the point P, lies on the required locus.

In the above given locus/equations of the point of intersection, the point $P=(0,4c)$ lies only on the equation depicted by option D.

Hence the required locus is

$y^2-4ax=16c^2$.