Prove that the area of the triangle formed by the tangents from the point $(x_{1}, y_{1})$ and the chord of contact is $(y_{1}^{2} - 4 a x_{1})^{\frac{3}{2}} \div 2 a$

Open in App

Solution

Tangents are drawn from $A (x_{1}, y_{1})$ and let the point of contact be $P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2})$

$P Q = \sqrt{{(a t_{2}^{2} - a t_{1}^{2})}^{2} + {(2 a t_{2} - 2 a t_{1})}^{2}} P Q = a \sqrt{{(t_{2} - t_{1})}^{2} {(t_{2} + t_{1})}^{2} + 4 {(t_{2} - t_{1})}^{2}} P Q = a (t_{2} - t_{1}) \sqrt{{(t_{2} + t_{1})}^{2} + 4} . . . . . . . . . . . (i)$

Equation of chord of contact w.r.t. $A$ is $T = 0$

$y y_{1} = 2 a (x + x_{1}) 2 a x - y y_{1} + 2 a x_{1} = 0...... (i i)$

Equation of chord joining $A B$ is

$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2} 2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0...... (i i i)$

Comparing $(i i)$ and $(i i i)$

$\frac{2}{2 a} = \frac{- (t_{1} + t_{2})}{- y_{1}} = \frac{2 a t_{1} t_{2}}{2 a x_{1}} \frac{1}{a} = \frac{t_{1} + t_{2}}{y_{1}} = \frac{t_{1} t_{2}}{x_{1}} \Rightarrow t_{1} + t_{2} = \frac{y_{1}}{a}, t_{1} t_{2} = \frac{x_{1}}{a} {(t_{1} + t_{2})}^{2} = \frac{y_{1}^{2}}{a^{2}} . . . . . . . . . . . (i v) {(t_{1} - t_{2})}^{2} = {(t_{1} + t_{2})}^{2} - 4 t_{1} t_{2} {(t_{1} - t_{2})}^{2} = \frac{y_{1}^{2}}{a^{2}} - \frac{{4 x}_{1}}{a} = \frac{y_{1}^{2} - 4 a x_{1}}{a^{2}} t_{2} - t_{1} = \frac{\sqrt{y_{1}^{2} - 4 a x_{1}}}{a} . . . . . . . . . (v)$

Substituting $(i v)$ and $(v)$ in $(i)$

$P Q = a \frac{\sqrt{y_{1}^{2} - 4 a x_{1}}}{a} \sqrt{\frac{y_{1}^{2}}{a^{2}} + 4} P Q = a \frac{\sqrt{y_{1}^{2} - 4 a x_{1}}}{a} \frac{\sqrt{y_{1}^{2} + 4 a^{2}}}{a} P Q = \frac{\sqrt{y_{1}^{2} - 4 a x_{1}} \sqrt{y_{1}^{2} + 4 a^{2}}}{a}$

Draw $A B$ perpendicular to $P Q$

$A B = \frac{| 2 a x_{1} - y_{1} (y_{1}) + 2 a x_{1} |}{\sqrt{{(2 a)}^{2} + {(- y_{1})}^{2}}} A B = \frac{{y_{1}}^{2} - 4 a x_{1}}{\sqrt{y_{1}^{2} + 4 a^{2}}}$

Area of $△ A P Q = \frac{1}{2} \times A B \times P Q$

$= \frac{1}{2} \times \frac{{y_{1}}^{2} - 4 a x_{1}}{\sqrt{y_{1}^{2} + 4 a^{2}}} \times \frac{\sqrt{y_{1}^{2} - 4 a x_{1}} \sqrt{y_{1}^{2} + 4 a^{2}}}{a} = \frac{{({y_{1}}^{2} - 4 a x_{1})}^{\frac{3}{2}}}{2 a} = {({y_{1}}^{2} - 4 a x_{1})}^{\frac{3}{2}} \div 2 a$

Hence proved.

Suggest Corrections

Similar questions

(x_{1}, y_{1})

\frac{\sqrt{y_{1}^{2} + 4 a^{2}} \sqrt{y_{1}^{2} - 4 a x_{1}}}{a} .

(3, 2)

x^{2} - 9 y^{2} = 9

(3, 2)

(x_{1}, y_{1}), (x_{2}, y_{2})

(x_{3}, y_{3})

y^{2} = 4 a x (a > 0)

\frac{1}{16 a} | (y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1}) |

(h, k)

x^{2} + y^{2} = a^{2}

(x_{1}, y_{1})

y^{2} = 4 a x

Join BYJU'S Learning Program