Prove that the locus of the point of intersection of tangents to the parabola `y^2=4ax` which meet at an angle `alpha` is `(x+a)^2tan^2a=y^2-4ax`.

Options

A. `(x+a)^(2)tan^(2)alpha=y^(2)-4ax`
B. `(x+a)tan^(2)alpha=y^(2)-4ax`
C. `(x-a)^(2)tan^(2)alpha=y^(2)-4ax`
D. none of these

Answer

A

Explanation

Let P(h, k) be the point of intersection of tangents to the parabola `y^(2)=4ax` is
`y=mh+a/m`
If it passes through (h, k), then
`k=mh+1/mrArrm^(2)h-mh+a=0`
Let `m_(1), m_(2)` be the roots of this equation. Then,
`m_(1)+m_(2)=k/h" and "m_(1)m_(2)=a/h`
Clearly, `m_(1), m_(2)` are the slopes of the tangents drawn from P.
`:." "tanalpha=(m_(1)-m_(2))/(1+m_(1)m_(2))=(sqrt((m_(1)+m_(2))^(2)-4m_(1)m_(2)))/(1+m_(1)m_(2))`
`rArr" "tanalpha(sqrt(k^(2)-4ah))/(a+h)rArrk^(2)-4ah=(a+h)^(2)tan^(2)alpha`
Hence, the locus of (h, k) is `y^(2)-4ax=(a+x)^(2)tan^(2)alpha`