Question

If a normal is drawn to the parabola $y^2 = 4x$ at point P(4, -4) which cut the parabola again at point Q, then the length of focal chord of the parabola which is parallel to PQ is equal to

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

5

Answer 2

The given parabola is $y^2 = 4x$, so $a=1$. The point P is (4, -4). The slope of the tangent at P is $m_t = \frac{2a}{y_1} = \frac{2(1)}{-4} = -\frac{1}{2}$. The slope of the normal at P is $m_n = -\frac{1}{m_t} = 2$. The equation of the normal at P(4, -4) is $y + 4 = 2(x - 4) \implies y = 2x - 12$. Substituting this into $y^2 = 4x$ gives $(2x - 12)^2 = 4x$, which simplifies to $x^2 - 13x + 36 = 0$. If $x_P = 4$, then $x_Q = 13 - 4 = 9$. The corresponding $y_Q = 2(9) - 12 = 6$. So, Q is (9, 6). The slope of PQ is $m_{PQ} = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$. A focal chord parallel to PQ has a slope $m=2$. The length of a focal chord of $y^2 = 4ax$ with slope $m$ is $L = \frac{4a(m^2+1)}{m^2}$. For $a=1$ and $m=2$, $L = \frac{4(1)(2^2+1)}{2^2} = \frac{4(5)}{4} = 5$.