Question

Tangents are drawn to $y^2 = 8x$ from point (−2,−3) which meet parabola at point 𝑃 and 𝑄. Then area of quadrilateral formed by tangents and normals drawn to parabola at points P and Q , is

• A. 125 / 2
• B. 125
• C. 65 / 2
• D. 65

Answer 1

Answer: (A)

125 / 2

Answer 2

The equation of the parabola is $y^2 = 8x$, which means $4a = 8$, so $a = 2$. The external point is $(x_1, y_1) = (-2, -3)$.

1. Chord of Contact: The equation of the chord of contact from $(-2, -3)$ to $y^2 = 8x$ is $yy_1 = 2a(x + x_1)$, which gives $y(-3) = 2(2)(x - 2)$, or $-3y = 4x - 8$, so $4x + 3y - 8 = 0$.

2. Points of Contact P and Q: To find P and Q, we solve the system of equations: $y^2 = 8x$ $4x + 3y - 8 = 0 \implies x = \frac{8 - 3y}{4}$ Substituting x into the parabola equation: $y^2 = 8 \left( \frac{8 - 3y}{4} \right)$ $y^2 = 2(8 - 3y)$ $y^2 = 16 - 6y$ $y^2 + 6y - 16 = 0$ $(y + 8)(y - 2) = 0$ So, the y-coordinates are $y = 2$ and $y = -8$. If $y = 2$, $x = \frac{8 - 3(2)}{4} = \frac{8 - 6}{4} = \frac{2}{4} = \frac{1}{2}$. So, $P = (\frac{1}{2}, 2)$. If $y = -8$, $x = \frac{8 - 3(-8)}{4} = \frac{8 + 24}{4} = \frac{32}{4} = 8$. So, $Q = (8, -8)$.

3. Equations of Tangents: Tangent at $P(\frac{1}{2}, 2)$: $y(2) = 2(2)(x + \frac{1}{2}) \implies 2y = 4x + 2 \implies 2x - y + 1 = 0$. Tangent at $Q(8, -8)$: $y(-8) = 2(2)(x + 8) \implies -8y = 4x + 32 \implies x + 2y + 8 = 0$.

4. Equations of Normals: The slope of the tangent at $(x_0, y_0)$ to $y^2 = 4ax$ is $\frac{2a}{y_0}$. Slope of tangent at $P(\frac{1}{2}, 2)$ is $\frac{2(2)}{2} = 2$. Slope of normal at P is $-\frac{1}{2}$. Normal at $P(\frac{1}{2}, 2)$: $y - 2 = -\frac{1}{2}(x - \frac{1}{2}) \implies 2y - 4 = -x + \frac{1}{2} \implies 2x + 4y - 9 = 0$. Slope of tangent at $Q(8, -8)$ is $\frac{2(2)}{-8} = -\frac{1}{2}$. Slope of normal at Q is $2$. Normal at $Q(8, -8)$: $y - (-8) = 2(x - 8) \implies y + 8 = 2x - 16 \implies 2x - y - 24 = 0$.

5. Quadrilateral Formation: The four lines are: $T_P: 2x - y + 1 = 0$ (slope 2) $T_Q: x + 2y + 8 = 0$ (slope -1/2) $N_P: 2x + 4y - 9 = 0$ (slope -1/2) $N_Q: 2x - y - 24 = 0$ (slope 2) Since $T_P \parallel N_Q$ and $T_Q \parallel N_P$, the quadrilateral is a parallelogram.

6. Area of the Parallelogram: The area of a parallelogram formed by the lines $a_1x+b_1y+c_1=0, a_1x+b_1y+d_1=0, a_2x+b_2y+c_2=0, a_2x+b_2y+d_2=0$ is given by $\frac{|(c_1-d_1)(c_2-d_2)|}{|a_1b_2 - a_2b_1|}$. Pair 1 (slopes 2): $2x - y + 1 = 0$ and $2x - y - 24 = 0$. Here $a_1=2, b_1=-1, c_1=1, d_1=-24$. Pair 2 (slopes -1/2): $x + 2y + 8 = 0$ and $2x + 4y - 9 = 0$. To match coefficients, we can write the second equation as $x + 2y - \frac{9}{2} = 0$. Here $a_2=1, b_2=2, c_2=8, d_2=-\frac{9}{2}$.

   Area $= \frac{|(1 - (-24))(8 - (-\frac{9}{2}))|}{|(2)(2) - (1)(-1)|} = \frac{|(25)(8 + \frac{9}{2})|}{|4 + 1|} = \frac{|25(\frac{16+9}{2})|}{5} = \frac{|25(\frac{25}{2})|}{5} = \frac{625/2}{5} = \frac{625}{10} = \frac{125}{2}$.