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Find the locus of the middle points of the chords of contact of orthogonal tangents to the parabola $\mathrm{y^2=4 a x.}$
Option: 1
$\mathrm{y^2=a(x+a)}$

Option: 2
$\mathrm{y^2=2 a(x-a)}$

Option: 3
$\mathrm{y^2=2 a(x+a)}$

Option: 4
$\mathrm{y^2=a(x-a)}$

Answers (1)

best_answer

The equation of the chord of contact of tangents from $\mathrm{(h, k)}$ is

$\mathrm{ \mathrm{yk}=2 \mathrm{a}(\mathrm{x}+\mathrm{h}) }$ $......(1)$

and the equation of chord of parabola whose middle point is $\mathrm{(\alpha, \beta)}$ is

$\mathrm{ y \beta-2 a(x+\alpha)=\beta^2-4 a \alpha }$ $......(2)$

Since (1) and (2) are same

$\mathrm{ \frac{k}{\beta}=\frac{2 a}{2 a}=\frac{2 a h}{2 a \alpha+\beta^2-4 a \alpha} }$

$\mathrm{\Rightarrow \mathrm{k}=\beta}$ and $\mathrm{\frac{2 \mathrm{ah}}{\beta^2-2 \mathrm{a} \alpha}=1 \Rightarrow \mathrm{h}=\frac{\beta^2-2 \mathrm{a} \alpha}{2 \mathrm{a}} \, \, and \, \, \mathrm{k}=\beta}$
Since the tangents from $(h, k)$ are at right angles, the point $(h, k)$ lies on the directrix

$\mathrm{ \begin{aligned} & x+a=0 \\ & \Rightarrow h+a=0 \Rightarrow \frac{\beta^2-2 a \alpha}{2 a}+a=0 \\ & \Rightarrow \beta^2-2 a \alpha+2 a^2=0 \Rightarrow \beta^2=2 a(\alpha-a) \end{aligned} }$

Hence locus of $\mathrm{(\alpha, \beta)}$ is

$\mathrm{ y^2=2 a(x-a) . }$

Posted by

HARSH KANKARIA

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