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Question

Answers

A. $x = 1$

B. $2x + 1 = 0$

C. $x = - 1$

D. $2x - 1 = 0$

Answer
Verified

Hint: Use the information that the locus of point p from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola. So, essentially, we need to find the directrix of the given parabola.

Complete step-by-step answer:

We have given the equation of parabola as ${y^2} = 4x$. Letâ€™s consider it as an equation â€¦â€¦..(1).

We know that the locus of point p from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola.

Also, the standard equation to the parabola is ${y^2} = 4ax$ and when we shift the origin to the point $(h,k)$ then this equation will become ${(y - k)^2} = 4p(x - h)$. It has focus $(h + p,k)$ and the directrix is $x = h - p$.

From equation (1), we get, $h = 0,k = 0,p = 1$. So, the directrix of equation (1) is $x = 0 - 1 = - 1$. Hence, required locus is $x = - 1$.

Note: Locus, as the word says, is the path of a point under given conditions. Here we observed that the path of the point will be the directrix of the given parabola then we solved the problem.

Complete step-by-step answer:

We have given the equation of parabola as ${y^2} = 4x$. Letâ€™s consider it as an equation â€¦â€¦..(1).

We know that the locus of point p from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola.

Also, the standard equation to the parabola is ${y^2} = 4ax$ and when we shift the origin to the point $(h,k)$ then this equation will become ${(y - k)^2} = 4p(x - h)$. It has focus $(h + p,k)$ and the directrix is $x = h - p$.

From equation (1), we get, $h = 0,k = 0,p = 1$. So, the directrix of equation (1) is $x = 0 - 1 = - 1$. Hence, required locus is $x = - 1$.

Note: Locus, as the word says, is the path of a point under given conditions. Here we observed that the path of the point will be the directrix of the given parabola then we solved the problem.

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