Question

PA and PB  are tangents drawn to y2=4ax from arbitrary point P . If the angle between tangents is π/4  then locus of point P  is​

Answer 1

Answer:

I hope this helps you.,.

Answer 2

Given:

$\mathrm{y}^{2}=4 \mathrm{ax}$

the angle between tangents is π/4 or 45°

Find:

locus of point P

Solution:

$\text { Let } P(\alpha, \beta) \text { be any point on the locus. Equation of pair of tangents from } P(\alpha, \beta) \text { to }$

$\text { the parabola } y^{2}=4 a x \text { is }$

$\begin{array}{l}{[\beta y-2 a(x+\alpha)]^{2}=\left(\beta^{2}-4 a \alpha\right)\left(y^{2}-4 a x\right)} \\\quad\left[\because T^{2}=S+S_{1}\right] \\\Rightarrow \beta^{2} y^{2}+4 a^{2}\left(x^{2}+\alpha^{2}+2 x \cdot \alpha\right)-4 a \beta y(x+\alpha) \\\Rightarrow \beta^{2} y^{2}-4 \beta^{2} a x-4 a \alpha y^{2}+16 a^{2} \alpha x \\=\beta^{2} y^{2}-4 \beta^{2} a x-4 a \alpha y^{2}+16 a^{2} \alpha x- \\4 a \beta x y-4 a \beta a y \ -(1)\end{array}$

$\text { Again, angle between the two of Eq. (i) is given as } 45^{\circ}$

$\begin{aligned}& \therefore \tan 45^{\circ}=\frac{2 \sqrt{h^{2}-a b}}{a+b} \\\Rightarrow & 1=\frac{2 \sqrt{h^{2}-a b}}{a+b} \Rightarrow a+b=2 \sqrt{h^{2}-a b}\end{aligned}$

$\begin{aligned}(a+b)^{2} &=4\left(h^{2}-a b\right) \\\left(4 a^{2}+4 a \alpha\right)^{2} &=4\left[4 a^{2} \beta^{2}-\left(4 a^{2}\right)(4 a \alpha)\right]\end{aligned}$

$\begin{array}{l}16 a^{2}(a+\alpha)^{2}=4 \cdot 4 a^{2}\left[\beta^{2}-4 a \alpha\right] \\\alpha^{2}+6 a \alpha+a^{2}-\beta^{2}=0 \\(\alpha+3 a)^{2}-\beta^{2}=8 a^{2}\end{array}$

$\text { Thus, the required equation of the locus is }(x+3 a)^{2}-y^{2}=8 a^{2} \text { which is a hyperbola. }$Hence locus of point P is​ hyperbola.