Question

Normals are drawn to the parabola 𝑦2 = 4𝑎𝑥 from any point on the line 𝑦 = 𝑏. Let A, B, C are feet of the normals, then the vertices of the triangle formed by tangents at A, B,C lie on

• A. 𝑥𝑦 = 𝑎2 +𝑏2
• B. 𝑥𝑦 = 𝑎2 − 𝑏2
• C. 𝑥𝑦 = −𝑎𝑏
• D. 𝑥𝑦 = 𝑎𝑏

Answer 1

Answer: (C)

𝑥𝑦 = −𝑎𝑏

Answer 2

Let the parabola be $y^2 = 4ax$. The equation of the normal to the parabola at the point $(at^2, 2at)$ is $tx + y - (2at + at^3) = 0$. If this normal passes through a point $P(h, b)$ on the line $y=b$, then $th + b - (2at + at^3) = 0$, which gives the cubic equation $at^3 + (2a - h)t - b = 0$. Let the roots of this equation be $t_1, t_2, t_3$. These roots correspond to the feet of the normals A, B, and C. From Vieta's formulas, we have $t_1 + t_2 + t_3 = 0$ and $t_1t_2t_3 = b/a$.

The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is $yt = x + at^2$. Consider the vertex formed by the intersection of the tangents at $t_i$ and $t_j$. The equations of these tangents are $yt_i = x + at_i^2$ and $yt_j = x + at_j^2$. Subtracting the second from the first gives $y(t_i - t_j) = a(t_i^2 - t_j^2) = a(t_i - t_j)(t_i + t_j)$. Assuming $t_i \neq t_j$, we get $y = a(t_i + t_j)$. Substituting this into the first tangent equation: $a(t_i + t_j)t_i = x + at_i^2$, which simplifies to $at_i^2 + at_it_j = x + at_i^2$, so $x = at_it_j$. Thus, any vertex of the triangle formed by the tangents has coordinates $(X, Y) = (at_it_j, a(t_i + t_j))$.

Since $t_1 + t_2 + t_3 = 0$, if $t_k$ is the root not involved in the pair $(t_i, t_j)$, then $t_k = -(t_i + t_j)$. Therefore, $Y = a(t_i + t_j)$ implies $t_i + t_j = Y/a$, and thus $t_k = -Y/a$. We have $X = at_it_j$. Using the product of roots $t_1t_2t_3 = b/a$, we can write $(at_it_j) \cdot t_k = b/a$. Substituting $X$ for $at_it_j$ and $-Y/a$ for $t_k$: $X \cdot (-Y/a) = b/a$ $-XY/a = b$ $XY = -ab$. Therefore, the vertices of the triangle formed by the tangents at A, B, and C lie on the curve $xy = -ab$.