14.2 Equation of chord of parabola whose midpoint is given

  Parabola

    1.0 Conic Section
    2.0 Parabola
    3.0 Standard equation of Parabola
    4.0 Focal distance of a point
    5.0 General equation of Parabola
    6.0 The generalized form of parabola: ${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$
    7.0 Parametric Co-ordinates
    8.0 Equation of tangent to a parabola
    9.0 Point of intersection of tangents at any two points on the parabola
    10.0 Equation of normal to the parabola
    11.0 Relation between parametric coefficients if normal meets parabola
    12.0 Important relations
    13.0 Circle through co-normal points
    14.0 Chord of contact

---

14.2 Equation of chord of parabola whose midpoint is given

When midpoint $M({x_1},{y_1})$ of chord $PQ$ given, the equation of chord can be find out using $T = {S_1}$ i.e., $$y{y_1} - 2a\left( {x + {x_1}} \right) = {y_1}^2 - 4a{x_1}$$

---

Question 9. Prove that the area of triangle formed by the tangents drawn from a point $P({x_1},{y_1})$ to the parabola ${y^2} = 4ax$ their chord of contact is $$\frac{{{{\left( {{y_1}^2 - 4a{x_1}} \right)}^{\frac{3}{2}}}}}{{2a}}$$

Solution: The equation of chord of contact $QR$ can be find out using $T=0$ i.e., $$y{y_1} = 2a\left( {x + {x_1}} \right)$$ or, $$y{y_1} - 2a\left( {x + {x_1}} \right) = 0$$

Let us assume the perpendicular from a point $P({x_1},{y_1})$ to the chord of contact be $PM$ and its length is

$$PM = \frac{{\left| {{y_1}{y_1} - 2a\left( {{x_1} + {x_1}} \right)} \right|}}{{\sqrt {{y_1}^2 + 4{a^2}} }}$$ $$ = \frac{{\left| {{y_1}^2 - 4a{x_1}} \right|}}{{\sqrt {{y_1}^2 + 4{a^2}} }}$$ and Length of chord of contact

$$QR = \frac{1}{{\left| a \right|}}\sqrt {({y_1}^2 - 4a{x_1})\left( {({y_1}^2 + 4{a^2}} \right)} $$

Now,

Area of $\Delta PQR = \frac{1}{2} \times QR \times PM$

$$ = \frac{1}{2}\frac{1}{{\left| a \right|}}\sqrt {({y_1}^2 - 4a{x_1})\left( {({y_1}^2 + 4{a^2}} \right)} \frac{{\left| {{y_1}^2 - 4a{x_1}} \right|}}{{\sqrt {{y_1}^2 + 4{a^2}} }}$$ $$ = \frac{{{{\left( {{y_1}^2 - 4a{x_1}} \right)}^{\frac{3}{2}}}}}{{2a}}$$